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જો ${C_r}$ એ $^n{C_r}$ દર્શાવે છે તો , $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$ મેળવો. (કે જ્યાં $n$ એ યુગ્મ પુર્ણાક છે )
$0$
${( - 1)^{n/2}}(n + 1)$
${( - 1)^n}(n + 2)$
${( - 1)^{n/2}}(n + 2)$
Solution
(d) We have $C_0^2 – 2C_1^2 + 3C_2^2 – ….. + {( – 1)^n}(n + 1)C_n^2$
$ = [C_0^2 – C_1^2 + C_2^2 – …. + {( – 1)^n}C_n^2] – [C_1^2 – 2C_2^2 + 3C_3^2…. – {( – 1)^n}nC_n^2]$
=${( – 1)^{n/2}}\frac{{n!}}{{(n/2)!(n/2)!}} – {( – 1)^{(n/2) – 1}}.\frac{1}{2}n\,{\,^n}{C_{n/2}}$
=${( – 1)^{n/2}}.\frac{{n!}}{{(n/2)!(n/2)!}}.\left( {1 + \frac{n}{2}} \right)$
$\therefore$ the value of the given expression is
$\frac{{2(n/2)\,!\,(n/2)!}}{{n!}} \times {( – 1)^{n/2}}.\frac{{(n)!}}{{(n/2)!(n/2)!}}\left( {1 + \frac{n}{2}} \right)$
$ = {( – 1)^{n/2}}(2 + n)$
