Some Special Series

It is given in the question that the n^th term of the series,

a_n = 3n² + 2n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From the above identities,

=

=

=

Hence, Sum of the series,

It is given in the question that the n^th term of the series,

a_n = n (n + 1) (n + 4)

Now, we need to find the sum of this series, S_n.

=

=

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From the above identities,

Hence, the Sum of the series,

It is given in the question that the n^th term of the series,

a_n = 4n³ + 6n² + 2n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From the above identities,

Hence, the Sum of the series,

It is given in the question that the n^th term of the series,

a_n = 3n² – 3n + 2

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

On simplifying,

Hence, the sum of the series,

It is given in the question that the n^th term of the series,

a_n = 2n² – 3n + 5

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

Hence, the sum of the series,

It is given in the question that the n^th term of the series,

a_n = n³ – 3ⁿ

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

→ (1)

The second term in the equation,,

forms a GP, with the common ratio, r = 3.

Sum of n terms of a GP, a, ar, ar², ar³…arⁿ.

Here, a= 3, r = 3;

So,

→ (2)

Substitute (2) in (1);

Hence, the sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is … 2² + 4² + 6² + 8² + … to n terms.

The series can be written as, [(2 x 1)², (2 x 2)²,

(2 x 3)²… (2 x n)²].

So, n^th term of the series,

a_n = (2n)² = 4n²

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is … 2³ + 4³ + 6³ + 8³ + … to n terms.

The series can be written as, [(2 x 1)³, (2 x 2)³,

(2 x 3)³… (2 x n)³].

So, n^th term of the series,

a_n = (2n)³ = 8n³

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is 5², 6², 7² …20².

The series can be written as, [(1 + 4)², (2 + 4)², (3 + 4)²… (16 + 4)²].

So, n^th term of the series,

a_n = (n + 4)²

With n = 16,

a_n = n² + 8n + 16

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

Here, n = 16 (from the question);

S_n = 2840

So, Sum of the series, S_n = 2840.

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + … to n terms.

The series can be written as, [(1 x (1 + 1)), (2 x (2 + 1)),

(3 x (3 + 1)),…(n x (n + 1))].

So, n^th term of the series,

a_n = n (n + 1)

a_n = n² + n

Now, we need to find the sum of this series, S_n.

Note:

V. Sum of first n natural numbers, 1 + 2 +3+…n,

VI. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

VII. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

VIII. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (3 × 8) + (6 × 11) + (9 × 14) + … to n terms.

The series can be written as, [(3 x 1) x (3 x 1 + 5)), (3 x 2) x (3 x 2 + 5))… (3n x (3n + 5))].

So, n^th term of the series,

a_n = 3n (3n + 5)

a_n = 9n² + 15n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2²) + (2 × 3²) + (3 × 4²) + … to n terms.

The series can be written as, [(1 x (1 + 1)²), (2 x (2 + 1)² … (n x (n + 1)²].

So, n^th term of the series,

a_n = n (n + 1)²

a_n = n³ + 2n² + n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2²) + (3 × 3²) + (5 × 4²) + … to n terms.

The series can be written as, [(1 x (1 + 1)²), (2 x (2 + 1)² … (2n-1 x (n + 1)²].

So, n^th term of the series,

a_n = (2n – 1) (n + 1)²

= (2n – 1) (n² + 2n + 1)

=2n³ + 3n² - 1

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (3 × 1²) + (5 × 2²) + (7 × 3²) + …to n terms.

The series can be written as, [(3 x 1²), (5 x 2² … ((2n + 1) x n²].

So, n^th term of the series,

a_n = (2n + 1) n²

a_n = 2n³ + n²

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2 x 3) + (2 × 3 x 4) + (3 × 4 x 5) + … to n terms.

The series can be written as, [(1 x (1 + 1) x (1 + 2)), (2 x (2 + 1) x (2 + 2) … (n x (n + 1) x (n + 2)].

So, n^th term of the series,

a_n = n (n + 1) (n + 2)

a_n = n³ + 3n² + 2n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2 × 4) + (2 × 3 × 7) + (3 × 4 × 10) + … to n terms.

The series can be written as, [(1 x (1 + 1) x (3 x 1 + 1)), (2 x (2 + 1) x (3 x 2 + 1))… (n x (n + 1) x (3 x n + 1)].

So, n^th term of the series,

a_n = n (n + 1) (3n + 1)

a_n = 3n³ + 4n² + n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is ….to n terms.

The series can be written as,….

So, n^th term of the series,

By the method of partial fractions, we can factorize the above term.

→ (1)

→ (2)

.

.

.

→ (n-1)^th equation

→ n^th equation

Now, we need to find the sum of this series, S_n.

This can be found out by adding the equation (1), (2)…up to n^th term.

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is … to n terms.

So, n^th term of the series,

By the method of partial fractions, we can factorize the above term.

On equating the like term on RHS and LHS,

2A + 2B = 0 → (a)

-A + B = 1 → (b)

On solving, we will get;

→ (1)

→ (2)

.

.

.

→ (n-1) ^th equation

→ n^th equation

Now, we need to find the sum of this series, S_n.

This can be found out by adding the equation (1), (2)…up to n^th term.

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (1 × 2²) + (2 × 3²) + (3 × 4²) + … to n terms.

The series can be written as, [(1 x (1 + 1)²), (2 x (2 + 1)² … (n x (n + 1)²].

So, n^th term of the series,

a_n = n (n + 1)²

a_n = n³ + 2n² + n

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, firs, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is to n terms

The series can be written as,

.

So, n^th term of the series,

The denominator of ‘a_n’ forms an AP with first term a = 1, last term = 2n-1 and common difference, d= 2.

Now, Sum of the AP,

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is 3, 15, 35, 63 … to n terms.

The series can be written as, [2² - 1, 4²-1, 6²-1… (2n)²-1].

So, n^th term of the series,

a_n = (2n)² - 1

a_n = 4n² - 1

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is 1 + 5 + 12 + 22 + 35 … to n terms.

This question can be solved by the method of difference.

Note:

Consider a sequence a₁, a₂, a₃ …such that the Sequence a₂ –a₁, a₃ – a₂… is either an. A.P. or a G.P.

The n^th term, of this sequence, is obtained as follows:

S = a₁ + a₂ + a₃ +…+ a_n–1 + a_n→ (1)

S = a₁ + a₂ +…+ a_n–2 + a_n–1 + a_n → (2)

Subtracting (2) from (1),

We get, a_n = a₁+ [(a₂–a₁) + (a₃–a₂) +… (a_n – a_n–1)].

Since the terms within the brackets are either in an A.P. or a G.P, we can find the value of a_n the n^th term.

Thus, we can find the sum of the n terms of the sequence as,

So,

By using the method of difference, we can find the n^th term of the expression.

S_n = 1 + 5 + 12 + 22 + 35 + ….. + a_n→ (1)

S_n = 1 + 5 + 12 + 22+ 35 + …. + a_n→ (2)

(1) – (2) → 0 = 1 + 4 + 7 + 10+ ….. - a_n

So, n^th term of the series,

a_n = 1 + 4 + 7 + 10 + ….

So, the n^th term form an AP, with the first term, a = 1; common difference, d = 3.

The required n^th term of the series is the same as the sum of n terms of AP.

Sum of n terms of an AP,

So, n^th term of the series,

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

In the given question we need to find the sum of the series.

For that, first, we need to find the n^th term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is 5 + 7 + 13 + 31 + 85 + …. + n terms.

This question can be solved by the method of difference.

Note:

Consider a sequence a₁, a₂, a₃ …such that the Sequence a₂ –a₁, a₃ – a₂… is either an. A.P. or a G.P.

The n^th term, of this sequence, is obtained as follows:

S = a₁ + a₂ + a₃ +…+ a_n–1 + a_n→ (1)

S = a₁ + a₂ +…+ a_n–2 + a_n–1 + a_n → (2)

Subtracting (2) from (1),

We get, a_n = a₁+ [(a₂–a₁) + (a₃–a₂) +… (a_n – a_n–1)].

Since the terms within the brackets are either in an A.P. or a G.P, we can find the value of a_n, the n^th term.

Thus, we can find the sum of the n terms of the sequence as,

So,

By using the method of difference, we can find the n^th term of the expression.

S_n = 5 + 7 + 13 + 31 + 85 + ….. + a_n→ (1)

S_n = 5 + 7 + 13 + 31+ 85 + …. + a_n→ (2)

(1) – (2) → 0 = 5 + 2 + 6 + 18 + 54+ …..+ (a_n – a_n-1) - a_n

So, n^th term of the series,

a_n = 5 + 2 + 6 + 18 + 54 + ….

In the resulting series obtained, starting from 2, 6, 18…forms a GP.

So, the n^th term forms a GP, with the first term, a = 2; common ratio, r = 3.

The required n^th term of the series is the same as the sum of n terms of GP and 5.

The GP is 2 + 6 + 18 + 54 +… (n-1) terms.

Sum of n terms of a GP,

Sum of (n-1) terms of a GP,

So, n^th term of the series,

Now, we need to find the sum of this series, S_n.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

→ (a)

The first term in (a) is a GP, with the first term, a = 3 and common ratio, r = 3.

Sum of n terms of GP,

So, Sum of the series,

Given,

To prove:

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So,

Now, the Left hand side of the condition given in the question can be written as,

The required LHS,

So,

_,

With,

Hence proved.

Given in the question, S_n denotes the sum of the cubes of the first n natural numbers.

s_n denotes the sum of the first n natural numbers.

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

III. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

IV. Sum of a constant k, N times,

So,

S_n denotes the sum of the cubes of the first n natural numbers. (Given data in the question).

s_n denotes the sum of the first n natural numbers.

To determine the given ratio in the question,

₌

So, the value of

It is required to find the sum of (2 + 4 + 6 + 8 +… 100).

Now, consider the series (2 + 4 + 6 + 8 +… 100).

If we take a common factor of 2 from all the terms, then,

the series becomes,

2 (1 + 2 + 3 + 4 +… 50).

So, we need to find the sum of first 50 natural numbers.

Note:

Sum of first n natural numbers, 1 + 2 +3+…n,

From the above identities,

So, Sum of first 50 natural numbers

(2 + 4 + 6 + 8 +… 100) = 2 (1 + 2 + 3 + 4 +… 50)

= 2 x 1275 = 2550

It is required to find the sum (41 + 42 + 43 + …. + 100).

(41 + 42 + 43 + …. + 100) = Sum of integers starting from 1 to 100 – Sum of integers starting from 1 to 40.

Note:

Sum of first n natural numbers, 1 + 2 +3+…n,

From the above identities,

So, Sum of integers starting from 1 to 100

So, Sum of integers starting from 1 to 40

(41 + 42 + 43 + …. + 100) = Sum of integers starting from 1 to 100 – Sum of integers starting from 1 to 40.

(41 + 42 + 43 + …. + 100) = 5050 – 820 = 4230

It is required to find the sum 11² + 12² + 13²+ …20²

11² + 12² + 13²+ …20² = Sum of squares of natural numbers starting from 1 to 20 – Sum of squares of natural numbers starting from 1 to 10.

Note:

Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

From the above identities,

Sum of squares of natural numbers starting from 1 to 20

Sum of squares of natural numbers starting from 1 to 10

11² + 12² + 13²+ …20²= Sum of squares of natural numbers starting from 1 to 20 – Sum of squares of natural numbers starting from 1 to 10.

11² + 12² + 13²+ …20² = 2870 – 385 = 2485

It is required to find the sum 6³ + 7³ + 8³ + 9³ + 10³.

6³ + 7³ + 8³ + 9³ + 10³= Sum of cubes of natural numbers starting from 1 to 10 – Sum of cubes of natural numbers starting from 1 to 5.

Note:

Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

From the above identities,

Sum of cubes of natural numbers starting from 1 to 10

Sum of cubes of natural numbers starting from 1 to 5

6³ + 7³ + 8³ + 9³ + 10³ = Sum of cubes of natural numbers starting from 1 to 10 – Sum of cubes of natural numbers starting from 1 to 5.

6³ + 7³ + 8³ + 9³ + 10³ = 3025 – 225 = 2800

It is given that,

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

From the above identities,

n² + n = 420

n² + n -420 = 0

(n – 20) (n + 21) = 0

n = 20 or -21

Since, n is the number of integers, n = 20

So,

It is given that,

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of cubes of first n natural numbers, 1³ + 2³ + 3³ +…..n³,

From the above identities,

We need to find,

We need to find the sum of the series {2² + 4² + 6² + …. + (2n)²}.

So, we can find it by using summation of the n^th term of the given series.

The n^th term of the series is (2n) ² = 4n²

(Given data)

a_n = 4n²

Now, sum of the series,

Note:

I. Sum of squares of first n natural numbers, 1² + 2² + 3²+….n²,

We need to find the sum of 10 terms of GP.

Sum of n terms of GP, with first term, a, common ratio, r,

So, the sum of given GP up to 10 terms, with a = √2,

r = √3, n = 10

The requires sum,

We need to find the sum of n terms of series whose r^th term is r + 2^r.

a_r = r + 2^r

So, n^th term, a_n = n + 2ⁿ

So, we can find the sum of the series by using summation of the n^th term of the given series.

→ (1)

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

Second term in (2) is a GP, with first term a = 2, common ratio r = 2.

Sum of n terms of GP, with the first term, a, common ratio, r,

So, the sum of given GP, with a = 2, r = 2

The required sum,

The sum of n terms of the series whose r^th term is (r + 2^r),