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If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =
$\frac{{n(n + 1)(n + 2)}}{{12}}$
$\frac{{n{{(n + 1)}^2}}}{{12}}$
$\frac{{n{{(n + 2)}^2}(n + 1)}}{{12}}$
None of these
Solution
(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 $n = 1, 2.$
