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 $\left(310\right)\cdot 2^{99}$

$S={}^n{C}_0\cdot a+{}^n{C}_1\cdot (a+d)+{}^n{C}_2\cdot (a+2d)+\cdots {+}^n{C}_n(a+nd)S=\frac{a+(a+nd)}{2}\cdot 2^n$

Given series is
$S_1=5C_0+8C_1+11C_2+\cdots$ utpo $101$ terms
Here $a=5,d=3,n=100$
$S_1=\frac{5+(5+3\cdot 100)}{2}\cdot 2^{100}\therefore S_1=310\cdot 2^{99}$


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