Find the sum of n terms of series whose mth term is 2^m + 2m

Given,

mth term = 2^m + 2m

1st term = 2¹ + 2 × 1

2nd term = 2² + 2 × 2

nth term = 2ⁿ + 2 × n

$S_n = [(2^1 + 2 \times 1) + (2^2 + 2 \times 2) +(2^3 + 2 \times 3)+…….. + (2^n + 2 \times n)] \\[1em] = [2^1 + 2^2 + 2^3 + ….. 2^n + (2 \times 1 + 2 \times 2 + 2 \times 3 + ….. + 2 \times n)] \\[1em] = (2^1 + 2^2 + 2^3 + ….. + 2^n) + 2(1 + 2 + 3 + ….. + n)……(1)$

Calculating :

2¹ + 2² + …….. + 2ⁿ

The above is an G.P. with a = 2 and r = 2.

By using formula,

S_n = $a\dfrac{r^n - 1}{r - 1}$

Substitute values, we get:

S_n = $2 \cdot \dfrac{2^n - 1}{2 - 1}$

= 2(2ⁿ - 1)

Calculating :

(1 + 2 + 3 + ….. + n)

The above is an A.P. with a = 1 and d = 1.

By using formula,

S_n = $\dfrac{n}{2}$[2a + (n - 1)d]

Substitute values, we get:

S_n = $\dfrac{n}{2}$ [2(1) + (n - 1)1]

= $\dfrac{n}{2}$ [2 + n - 1]

= $\dfrac{n}{2}$(n + 1).

Substitute values in (1) we get,

S_n = 2(2ⁿ - 1) + $2 \times \dfrac{n}{2}$(n + 1)

= 2(2ⁿ - 1) + n(n + 1)

Hence, S_n = 2(2ⁿ - 1) + n(n + 1).