Question

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

• A. $0$
• B. $\left(305\right)\cdot 2^{100}$
• C. $\left(305\right)\cdot 2^{99}$
• D. $1$

Answer 1

The correct option is A $0$

Let $S=5C_0-8C_1+11C_2-\dots$ upto $101$ terms
$=\sum _{r=0}^{100}(-1{)}^r\cdot (3r+5)\cdot C_r=\sum _{r=0}^{100}(-1{)}^r\cdot (3r+5)\cdot {}^{100}{C}_r=\sum _{r=0}^{100}(-1{)}^r\cdot 3r\cdot {}^{100}{C}_r+\sum _{r=0}^{100}(-1{)}^r\cdot 5\cdot {}^{100}{C}_r=3\sum _{r=0}^{100}(-1{)}^r\cdot r\cdot {}^{100}{C}_r+5\sum _{r=0}^{100}(-1{)}^r\cdot {}^{100}{C}_r=3\sum _{r=1}^{100}(-1{)}^r\cdot 100\cdot {}^{99}{C}_{r-1}+5[{}^{100}{C}_0-{}^{100}{C}_1+{}^{100}{C}_2-\dots +{}^{100}{C}_{100}]=-300\sum _{r=1}^{100}{}^{99}{C}_{r-1}(-1{)}^{r-1}+5(0)=0$



Alternate solution:
$S=5C_0-8C_1+11C_2-\dots +305C_{100}S=305C_{100}-302C_{99}+\dots +5C_0\Rightarrow 2S=310[C_0-C_1+C_2-\dots +C_{100}]\Rightarrow 2S=310\times \left[0\right]\therefore S=0$