Let-A-1-2-3-n-if-a-i-is-the-minimum-element-of-the-set-A-where-A-denotes-the-subset-of-A-containing-exactly-three-elements-and-X-denotes-the-set-of-A-i-s-then-evaluate-A-i-X-

Question Number 22044 by Tinkutara last updated on 10/Oct/17

Let A = {1, 2, 3, \dots . ., n}, if a_{i} is the

minimum element of the set A; (where

A; denotes the subset of A containing

exactly three elements) and X denotes

the set of A_{i}^{'} s, then evaluate \sum_{A_{i} \in X} a .

Answered by revenge last updated on 17/Oct/17

R e q u i r e d s u m = 1 \cdot^{n - 1} C_{2} + 2 \cdot^{n - 2} C_{2} + 3 \cdot^{n - 3} C_{2} + \dots + (n - 2) \cdot^{2} C_{2}

I t i s e q u a l t o t h e c o e f f i c i e n t o f x^{2} i n t h e e x p a n s i o n o f

{(1 + x)}^{n - 1} + 2 {(1 + x)}^{n - 2} + 3 {(1 + x)}^{n - 3} + \dots + (n - 2) {(1 + x)}^{2}

S = {(1 + x)}^{n - 1} + 2 {(1 + x)}^{n - 2} + 3 {(1 + x)}^{n - 3} + \dots + (n - 2) {(1 + x)}^{2}

\frac{S}{1 + x} = {(1 + x)}^{n - 2} + 2 {(1 + x)}^{n - 3} + \dots + (n - 3) {(1 + x)}^{2} + (n - 2) (1 + x)

S - \frac{S}{1 + x} = {(1 + x)}^{n - 1} + {(1 + x)}^{n - 2} + {(1 + x)}^{n - 3} + \dots + {(1 + x)}^{2} - (n - 2) (1 + x)

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

S = \frac{{(1 + x)}^{3} [{(1 + x)}^{n - 2} - 1]}{x^{2}} - \frac{(n - 2) {(1 + x)}^{2}}{x}

S = \frac{{(1 + x)}^{n + 1}}{x^{2}} - \frac{{(1 + x)}^{3}}{x^{2}} - (n - 2) \frac{1 + 2 x + x^{2}}{x}

x^{2} t e r m c a n b e o n l y o b t a i n e d f r o m \frac{{(1 + x)}^{n + 1}}{x^{2}} . S o w e r e q u i r e

x^{4} i n {(1 + x)}^{n + 1}, w h i c h h a s t h e c o e f f i c i e n t^{n + 1} C_{4} .

Commented by Tinkutara last updated on 17/Oct/17

Thank you very much Sir!