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JEE Mains · Maths · STD 11 - 4.1 complex nubers

माना \(i =\sqrt{-1}\) है यदि \(\frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}=k\) है, तथा \(n =[| k |],| k |\) का महत्तम पूर्णांक भाग है, तो \(\sum_{j=0}^{n+5}(j+5)^{2}-\sum_{j=0}^{n+5}(j+5)\) बराबर है 

  1. A \(620\)
  2. B \(310\)
  3. C \(155\)
  4. D \(280\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(310\)

Step-by-step Solution

Detailed explanation

\(K=\frac{1}{2^{9}}\left[\frac{\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{21}}{\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i \right)^{24}}+\frac{\left(\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{21}}{\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i \right)^{24}}\right]\)…
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