If n is a positive integer and {C_k} = {,^n}{ | vedclass

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Question

(d) $\sum\limits_{k = 1}^n {{k^3}} {\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2} = \sum\limits_{k = 1}^n {{k^3}} {\left( {\frac{{n - k + 1}}{k}}

Vedclass · Accepted Answer

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Vedclass · Answer

(d) $\sum\limits_{k = 1}^n {{k^3}} {\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2} = \sum\limits_{k = 1}^n {{k^3}} {\left( {\frac{{n - k + 1}}{k}} \right)^2}$

$\sum\limits_{k = 1}^n k {(n - k + 1)^2} = \sum\limits_{k = 1}^n {k[{{(n + 1)}^2} - 2k(n + 1) + {k^2}]} $

$ = {(n + 1)^2}\sum\limits_{k = 1}^n {k - 2(n + 1)\sum\limits_{k = 1}^n {{k^2} + \sum\limits_{k = 1}^n {{k^3}} } } $

$ = {(n + 1)^2}.\frac{{n(n + 1)}}{2} - 2(n + 1).\frac{{n(n + 1)(2n + 1)}}{6}$$ + \frac{{{n^2}{{(n + 1)}^2}}}{4}$

$ = \frac{{n{{(n + 1)}^2}}}{{12}}[6(n + 1) - 4(2n + 1) + 3n$]

$ = \frac{{n{{(n + 1)}^2}}}{{12}}.(n + 2) = \frac{{n(n + 2){{(n + 1)}^2}}}{{12}}$

Trick : Check by taking $n = 1, 2.$

If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =

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