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

माना \(a_1=b_1=1\) तथा \(a_n=a_{n-1}+(n-1)\), \(\mathrm{b}_{\mathrm{n}}=\mathrm{b}_{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-1}, \forall \mathrm{n} \geq 2\) है। यदि \(\mathrm{S}=\sum_{\mathrm{n}=1}^{10} \frac{\mathrm{b}_{\mathrm{n}}}{2^{\mathrm{n}}}\) तथा \(\mathrm{T}=\sum_{\mathrm{n}=1}^8 \frac{\mathrm{n}}{2^{\mathrm{n}-1}}\), है तो \(2^7(2 \mathrm{~S}-\mathrm{T})\) बराबर है_______. 

  1. A \(461\)
  2. B \(460\)
  3. C \(462\)
  4. D \(465\)
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Answer & Solution

Correct Answer

(A) \(461\)

Step-by-step Solution

Detailed explanation

\(\text { As, } S=\frac{b_1}{2}+\frac{b_2}{2^2}+\ldots \ldots .+\frac{b_9}{2^9}+\frac{b_{10}}{2^{10}}\) \(\Rightarrow \frac{S}{2}=\quad \frac{b_1}{2^2}+\frac{b_2}{2^3}+\ldots \ldots+\frac{b_9}{2^{10}}+\frac{b_{10}}{2^{11}}\) \(\text { subtracting }\)…
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