Question

Let $S_k = 3^k ·{}^{ 100}C_0 ·{}^{ 100}C_k - 3^{k - 1}· {}^{100}C_1 · {}^{99}C_{k - 1}+ 3^{k - 2} ·{}^{100}C_2 · {}^{98}C_{k - 2} ...... + (- 1)^{k} · {}^{100}C_k · {}^{100 - k}C_0 ,$ then value of $\displaystyle\sum_{ r =0}^{100} S _{ r } S _{100- r }$ 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)

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