Question-59639

Question Number 59639 by Tawa1 last updated on 12/May/19

Commented by Tawa1 last updated on 12/May/19

Please help me to continue . My final answer is wrong .

I got : S = \frac{{2 n}^{3} + {3 n}^{2} + n}{6} but answer is S = \frac{n}{6} ({14 n}^{2} - 9 n + 1)

Commented by tanmay last updated on 12/May/19

T_{r} = {(r \times \frac{n_{c_{r}}}{n_{c_{r - 1}}})}^{2} = {(\frac{r \times \frac{n!}{r! (n - r)!}}{\frac{n!}{(n - r + 1)! (r - 1)!}})}^{2}

= {(\frac{r \times (r - 1)! (n - r + 1)!}{r! (n - r)!})}^{2}

= {(n - r + 1)}^{2}

T_{r} = {(n + 1)}^{2} - 2 (n + 1) r + r^{2}

T_{1} = {(n + 1)}^{2} - 2 (n + 1) \times 1 + 1^{2}

T_{2} = {(n + 1)}^{2} - 2 (n + 1) \times 2 + 2^{2}

T_{3} = {(n + 1)}^{2} - 2 (n + 1) \times 3 + 3^{2}

\dots .

\dots .

T_{n} = {(n + 1)}^{2} - 2 (n + 1) \times n + n^{2}

n o w a d d t h e m

S = n \times {(n + 1)}^{2} - 2 (n + 1) (1 + 2 + 3 + . . + n) + (1^{2} + 2^{2} + . . + n^{2})

S = n {(n + 1)}^{2} - 2 (n + 1) \times \frac{n (n + 1)}{2} + \frac{n (n + 1) (2 n + 1)}{6}

S = \frac{n (2 n^{2} + 3 n + 1)}{6}

S = \frac{2 n^{3} + 3 n^{2} + n}{6}

Commented by Tawa1 last updated on 13/May/19

That means am right sir . God bless you

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