Question

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

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

Answer 1

The correct option is C $2^{100}$

$S_n=1+q+q^2+\cdots +q^n\Rightarrow S_n=\frac{1-q^{n+1}}{1-q}\cdots \left(1\right)$

$T_n=1+\left(\frac{q+1}{2}\right)+\left(\frac{q+1}{2}\right)+\cdots +{\left(\frac{q-1}{2}\right)}^n\Rightarrow T_n=\frac{1-{\left(\frac{q+1}{2}\right)}^{n+1}}{1-\left(\frac{q+1}{2}\right)}\Rightarrow T_n=\frac{2^{n+1}-(q+1{)}^{n+1}}{2^n.(1-q)}\Rightarrow T_{100}=\frac{2^{101}-(q+1{)}^{101}}{2^{100}(1-q)}\cdots \left(2\right)$

Now, ${}^{101}{C}_1+{}^{101}{C}_2\cdot S_1+\cdots +{}^{101}{C}_{101}\cdot S_{100}=\alpha T_{100}\Rightarrow \sum _{r=1}^{101}{}^{101}{C}_r{S}_{r-1}=\alpha T_{100}$
From equation $\left(1\right)$
$S_{r-1}=\frac{1-q^r}{1-q}$
$\Rightarrow \sum _{r=1}^{101}{}^{101}{C}_r\left(\frac{1-q^r}{1-q}\right)=\alpha T_{100}$
$\Rightarrow \frac{1}{1-q}[\sum _{r=1}^{101}{}^{101}{C}_r-\sum _{r=1}^{101}{}^{101}{C}_r\cdot q^r]=\alpha T_{100}\Rightarrow \frac{1}{1-q}\left[\right(2^{101}-1)-((1+q{)}^{101}-1)]=\alpha T_{100}$

From equation $\left(2\right)$, we get
$\frac{1}{1-q}[2^{101}-(1+q{)}^{101}]=\alpha \cdot \left[\frac{2^{101}-(1+q{)}^{101}}{2^{100}(1-q)}\right]\Rightarrow \alpha =2^{100}$