Question

If $C_r$ represents ${}^{100}{C}_r,$ then $5C_0+8C_1+11C_2+\dots$ upto $101$ terms is equal to

• A. $\left(305\right)\cdot 2^{100}$
• B. $\left(305\right)\cdot 2^{99}$
• C. $\left(310\right)\cdot 2^{100}$
• D. $\left(310\right)\cdot 2^{99}$

Answer 1

The correct option is D $\left(310\right)\cdot 2^{99}$

Let $S=5C_0+8C_1+11C_2+\dots$ upto $101$ terms
$=\sum _{r=0}^{100}(3r+5)\cdot C_r=\sum _{r=0}^{100}(3r+5)\cdot {}^{100}{C}_r=\sum _{r=0}^{100}3r\cdot {}^{100}{C}_r+\sum _{r=0}^{100}5\cdot {}^{100}{C}_r=3\sum _{r=0}^{100}r\cdot {}^{100}{C}_r+5\sum _{r=0}^{100}{}^{100}{C}_r=3\sum _{r=1}^{100}100\cdot {}^{99}{C}_{r-1}+5\cdot 2^{100}=300\cdot 2^{99}+5\cdot 2^{100}=310\cdot 2^{99}$


Alternate solution:
$S=5C_0+8C_1+11C_2+\dots +305C_{100}S=305C_{100}+302C_{99}+\dots +5C_0\Rightarrow 2S=310[C_0+C_1+C_2+\dots +C_{100}]\Rightarrow 2S=310\cdot 2^{100}\therefore S=310\cdot 2^{99}$