Question

Sum of the series ${\left({}^{100}{C}_1\right)}^2+2{\left({}^{100}{C}_2\right)}^2+3{\left({}^{100}{C}_3\right)}^2+\dots \dots ..+100{\left({}^{100}{C}_{100}\right)}^2$ equals

• A. $\frac{2^{99}[1\cdot 3\cdot 5\dots \dots ..(199)]}{(99)!}$
• B. $100\cdot {}^{200}{C}_{100}$
• C. $50\cdot {}^{200}{C}_{100}$
• D. $100\cdot {}^{199}{C}_{99}$

Answer 1

Answer: (D)

$S=0\cdot {\left({}^{100}{C}_0\right)}^2+1\cdot {\left({}^{100}{C}_1\right)}^2+2\cdot {\left({}^{100}{C}_2\right)}^2+3\cdot {\left({}^{100}{C}_3\right)}^2+\dots \dots \dots +100\cdot {\left({}^{100}{C}_{100}\right)}^2$
$S=100\cdot {\left({}^{100}{C}_0\right)}^2+99\cdot {\left({}^{100}{C}_1\right)}^2+\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots +0.\dots \dots \dots +{\left({}^{100}{C}_{100}\right)}^2$
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$2S=100\left[{\left({}^{100}{C}_0\right)}^2+{\left({}^{100}{C}_1\right)}^2+{\left({}^{100}{C}_2\right)}^2+\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots +{\left({}^{100}{C}_{100}\right)}^2\right]$
$2S=100\cdot {}^{200}{C}_{100}=100\cdot \frac{(200)!}{100!\cdot 100!}\Rightarrow 2S=\frac{100\cdot 200\cdot (199)!}{100\cdot (99)!\cdot (100)!}\Rightarrow S=100\cdot {}^{199}{C}_{99}$