Question

Let $S_n+1+q+q^2+.......+q^n$ and $T_n=1+\left(\frac{q+1}{2}\right)+{\left(\frac{q+1}{2}\right)}^2+.....+{\left(\frac{q+1}{2}\right)}^n$ where q is a real number and $q\ne 1$.
If ${}^{101}{C}_1+{}^{101}{C}_2.S_1+......+{}^{101}{C}_{101}.S_{100}=\alpha T_{100}$, then $\alpha$ is equal to :-

• A. $2^{100}$
• B. 200
• C. $2^{99}$
• D. 202

Answer 1

Answer: (A)

${}^{101}{C}_1{+}^{101}{C}_2{S}_1+.....{+}^{101}{C}_{101}{S}_{100}=\alpha T_{100}$
${}^{101}{C}_1{+}^{101}{C}_2\left(1+q\right){+}^{101}{C}_3\left(1+q+q^2\right)+....{+}^{101}{C}_{101}\left(1+q+....+q^{100}\right)$
$=2\alpha \frac{\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)}{\left(1-q\right)}$
${\Rightarrow }^{101}{C}_1\left(1-q\right){+}^{101}{C}_2\left(1-q^2\right)+.....{+}^{101}{C}_{101}\left(1-q^{101}\right)$
$=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)$
$\Rightarrow \left(2^{101}-1\right)-\left({\left(1+q\right)}^{101}-1\right)$
$=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)$
$\Rightarrow 2^{101}\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)$
$\Rightarrow \alpha =2^{100}$