Question

For any \(n≥0\) the value of \(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\) is 

• A. \(2nC_{n-1}\)
• B. \(8nC_n\)
• C. \(2nC_{n+1}\)
• D. \(nC_{n-2}\)
• E. \(2nC_n\)

Answer 1

Answer: (E)

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)