Question

Let $S_k=3^k\cdot {}^{100}{C}_0\cdot {}^{100}{C}_k-3^{k-1}\cdot {}^{100}{C}_1\cdot {}^{99}{C}_{k-1}+3^{k-2}\cdot {}^{100}{C}_2\cdot {}^{98}{C}_{k-2}......+(-1{)}^k\cdot {}^{100}{C}_k\cdot {}^{100-k}{C}_0,$ then value of $\sum _{r=0}^{100}{S}_r{S}_{100-r}$ is equal to

• A. ${}^{200}{C}_{100}$
• B. $2^{100}\cdot {}^{200}{C}_{100}$
• C. $2^{200}\cdot {}^{200}{C}_{100}$
• D. $2^{99}\cdot {}^{200}{C}_{99}$

Answer 1

Answer: (B)

$S_k=\sum _{r=0}^k(-1{)}^r{}^{100}{C}_r\cdot {}^{100}{C}_{k-r}\cdot 3^{k-r}=\sum _{r=0}^k(-1{)}^r\frac{100!}{r!(100-r)!}\cdot \frac{(100-r)!}{(k-r)!\cdot (100-k)!}\cdot 3^{k-r}$
$S_k=\sum _{r=0}^k(-1{)}^r\cdot \frac{100!}{k!\cdot (100-k)!}\cdot \frac{k!}{r!\cdot (100-r)!}\cdot 3^{k-r}=\sum _{r=0}^k(-1{)}^r\cdot {}^{100}{C}_k\cdot {}^k{C}_r\cdot 3^{k-r}$
$S_k=\sum _{r=0}^k{}^{100}{C}_k\cdot (2{)}^k$
$\sum _{r=0}^{100}{S}_r{S}_{100-r}=\sum _{r=0}^{100}{}^{100}{C}_r\cdot 2^r\times {}^{100}{C}_{100-r}\cdot 2^{100-r}$
$=2^{100}\sum _{r=0}^{100}{\left({}^{100}{C}_r\right)}^2=2^{100\cdot {}^{200}{C}_{100}\cdot }$