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JEE Mains · Maths · STD 11 - 8. sequence and series

माना \(S _{ n }=1+ q + q ^{2}+\ldots \ldots+ q ^{ n }\) तथा \(T _{ n }=1+\left(\frac{ q +1}{2}\right)+\left(\frac{ q +1}{2}\right)^{2}+\ldots \ldots .+\left(\frac{ q +1}{2}\right)^{ n }\) जहाँ \(q\) एक वास्तविक संख्या है तथा \(q \neq 1\) । यदि \({ }^{101} C _{1}+{ }^{101} C _{2} . S _{1}+\ldots \ldots+{ }^{101} C _{101} . S _{100}=\alpha T _{100}\) तो \(\alpha\) बराबर है 

  1. A \(2^{99}\)
  2. B \(202\)
  3. C \(200\)
  4. D \(2^{100}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2^{100}\)

Step-by-step Solution

Detailed explanation

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