Set Sn=1+q+q2+…+qn and Tn=1+q+12+q+122+…+q+12nwhere q is a real number and q≠1 If 101C1+101C2S1+…+101 C101S100=αT100 , then α is equal to

# Set ${S}_{n}=1+q+{q}^{2}+\dots +{q}^{n}$ and ${T}_{n}=1+\left(\frac{q+1}{2}\right)+{\left(\frac{q+1}{2}\right)}^{2}+\dots +{\left(\frac{q+1}{2}\right)}^{n}$where $q$ is a real number and $q\ne 1$  ${C}_{101}{S}_{100}=\alpha {T}_{100}$ , then $\alpha$ is equal to

1. A

${2}^{100}$

2. B

${2}^{99}$

3. C

202

4. D

200

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### Solution:

Here, ${C}_{101}{S}_{100}=\alpha {T}_{100}$

${⇒}^{101}{C}_{1}{+}^{101}{C}_{2}\left(1+q\right){+}^{101}{C}_{3}\left(1+q+{q}^{2}\right)+\dots$${+}^{101}{C}_{101}\left(1+q+\dots +{q}^{100}\right)$

$\begin{array}{l}=2\alpha \frac{\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)}{\left(1-q\right)}\\ {⇒}^{101}{C}_{1}\left(1-q\right){+}^{101}{C}_{2}\left(1-{q}^{2}\right)+\dots {+}^{101}{C}_{101}\left(1-{q}^{101}\right)\\ =2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)\\ ⇒\left({2}^{101}-1\right)-\left[\left(1+q{\right)}^{101}-1\right]=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)\\ ⇒{2}^{101}\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)⇒\alpha ={2}^{100}\end{array}$

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