Question

If S₁, S₂, ..., S_n are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S₁ + S₂ + 2S₃ + 3S₄ + ... (n − 1) S_n = 1ⁿ + 2ⁿ + 3ⁿ + ... + nⁿ.

Answer 1

Given:

\[ S_1 , S_2 , . . . , S_n\text {  are the sum of n terms of an G . P . whose first term is 1 in each case and the common ratios are } 1, 2, 3, . . . , n . \]

\[ \therefore S_1 = 1 + 1 + 1 + . . .\text {  n terms } = n . . . \left( 1 \right)\]

\[ S_2 = \frac{1\left( 2^n - 1 \right)}{2 - 1} = 2^n - 1 . . . \left( 2 \right)\]

\[ S_3 = \frac{1\left( 3^n - 1 \right)}{3 - 1} = \frac{3^n - 1}{2} . . . \left( 3 \right)\]

\[ S_4 = \frac{1\left( 4^n - 1 \right)}{4 - 1} = \frac{4^n - 1}{3} . . . \left( 4 \right)\]

\[ S_n = \frac{1\left( n^n - 1 \right)}{n - 1} = \frac{n^n - 1}{n - 1} . . . . . . . . . . . . \left( n \right)\]

\[\text { Now, LHS } = S_1 + S_2 + 2 S_3 + 3 S_4 + . . . + \left( n - 1 \right) S_n \]

\[ = n + 2^n - 1 + 3^n - 1 + 4^n - 1 + . . . + n^n - 1 \left[ \text { Using } \left( 1 \right), \left( 2 \right), \left( 3 \right), . . . , \left( n \right) \right]\]

\[ = n + \left( 2^n + 3^n + 4^n + . . . + n^n \right) - \left[ 1 + 1 + 1 + . . . + \left( n - 1 \right) \text { times } \right]\]

\[ = n + \left( 2^n + 3^n + 4^n + . . . + n^n \right) - \left( n - 1 \right)\]

\[ = n + \left( 2^n + 3^n + 4^n + . . . + n^n \right) - n + 1\]

\[ = 1 + 2^n + 3^n + 4^n + . . . + n^n \]

\[ = 1^n + 2^n + 3^n + 4^n + . . . + n^n \]

= RHS

Hence proved .