Let S_n = 1 + q + q^2 + ..... + q^n and T_n = 1 + ( q + 1 2 ) + (… - Vidyadip

MCQ

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
$2^{99}$
B
$202$
C
$200$
✓
$2^{100}$

✓

Answer

Correct option: D.

$2^{100}$

d
$\sum\limits_{r = 1}^{101} {{\,^{101}}{C_r}{s_{r - 1}}} $

$ = \sum\limits_{r = 1}^{101} {{\,^{101}}{C_r}\frac{{{q^r} - 1}}{{q - 1}}} $

$ = \frac{1}{{q - 1}}\left( {\sum\limits_{r = 1}^{101} {{\,^{101}}{C_r}{q^r} - \sum\limits_{r = 1}^{101} {{\,^{101}}{C_r}} } } \right)$

$ = \frac{1}{{q - 1}}\left( {{{\left( {1 + q} \right)}^{101}} - 1 - {2^{101}} + 1} \right)$

$ = \frac{\alpha }{{{2^{100}}}}\left( {\frac{{{{\left( {1 + q} \right)}^{101}} - {2^{101}}}}{{q - 1}}} \right)$

$ \Rightarrow \alpha = {2^{100}}$

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