Question

Sum to $n$ term the following series $2 \times 1 + 3 \times 2 + 4 \times 4 + 5 \times 8 + \dots$

Accepted Answer

The given series $2 \times 1 + 3 \times 2 + 4 \times 4 + 5 \times 8 + \dots$ is arithmetic $-$ geometric series each of whose term is obtained by multiplying the corresponding terms of $A . P .$ $2, 3, 4, 5, \dots$ and $G . P .$ $1, 2, 4, 8, \dots$

Let $S_{n} = 2 \times 1 + 3 \times 2 + 4 \times 4 + 5 \times 8 + \dots . + n terms$

We know that, the nth term of a $A . G . P .$ series is $T_{n} = [a + (n - 1) d] r^{n - 1}$ .

Here, $n^{t h}$ term of arithmetic geometric series be $[2 + (n - 1) \cdot 1] [1 \cdot 2^{n - 1}] = (n + 1) (2^{n - 1})$

$\Rightarrow S_{n} = 2 \times 1 + 3 \times 2 + 4 \times 4 + 5 \times 8 + \dots . + n 2^{n - 2} + (n + 1) 2^{n - 1}$ (i)

Multiplying both sides with the common ratio of $G . P .$ , $r = 2$

$∴$ $2 S_{n} = 2 \times 2 + 3 \times 4 + 4 \times 8 + \dots + \dots . + n 2^{n - 1} + (n + 1) 2^{n}$ (ii)

Now, equation (i) $-$ equation (ii) gives

$- S_{n} = 2 \times 1 + 1 \times 2 + 1 \times 4 + 1 \times 8 \dots + 2^{n - 1} - (n + 1) 2^{n}$

We know that, the sum of finite terms of a $G . P .$ series is $S_{n} = \frac{a (r^{n} - 1)}{r - 1}, r > 1$ .

Here, Series on R.H.S. be an finite series (leaving first and last term) with $a = 2, r = 2$

$= 2 + \frac{2 [2^{n - 1} - 1]}{2 - 1} - (n + 1) 2^{n}$

$\Rightarrow - S_{n} = 2 + \frac{2 [2^{n - 1} - 1]}{2 - 1} - (n + 1) 2^{n}$

$\Rightarrow - S_{n} = 2 + 2^{n} - 2 - (n + 1) 2^{n} = [1 - n - 1] 2^{n}$

$\Rightarrow S_{n} = n \cdot 2^{n}$

$∴$ Sum to finite of the given series $S_{n} = n \cdot 2^{n}$ .

M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 8: EXERCISE 10.8

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 8: EXERCISE 10.8

Questions from M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 8: EXERCISE 10.8 with Hints & Solutions

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