Question:
For any \(n≥0\) the value of \(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\) is
For any \(n≥0\) the value of \(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\) is
Updated On: Jun 3, 2024
\(2nC_{n-1}\)
\(8nC_n\)
\(2nC_{n+1}\)
\(nC_{n-2}\)
\(2nC_n\)
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The Correct Option is
Solution and Explanation
Given that
\(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\)
condition is \(n≥0\)
So lets put \(n=1\)
Then the parent term will be
\(\dfrac{3+7}{5}=2\)
So we can represent the term as : \(2nC_n\) (_Ans)
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