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: 

\(\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}\)

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 : \(2nC_n\)

Answer 2

Let \( S = \sum_{r=0}^n \frac{(4r+3) \binom{n}{r}^2}{2n+3} \). 

Using the identity:

\[ \sum_{r=0}^n r \binom{n}{r}^2 = n \binom{2n-1}{n-1} \]

and

\[ \sum_{r=0}^n \binom{n}{r}^2 = \binom{2n}{n} \]

We can rewrite the sum as:

\[ S = \sum_{r=0}^n \frac{4r \binom{n}{r}^2}{2n+3} + \sum_{r=0}^n \frac{3 \binom{n}{r}^2}{2n+3} \] \[ S = \frac{4}{2n+3} \sum_{r=0}^n r \binom{n}{r}^2 + \frac{3}{2n+3} \sum_{r=0}^n \binom{n}{r}^2 \]

Using the identities above:

\[ S = \frac{4}{2n+3} n \binom{2n-1}{n-1} + \frac{3}{2n+3} \binom{2n}{n} \] \[ S = \frac{4n \binom{2n-1}{n-1} + 3 \binom{2n}{n}}{2n+3} \]

We can use the identity \( \binom{n}{r} = \binom{n}{n-r} \):

\[ \sum_{r=0}^n r \binom{n}{r}^2 = \sum_{r=0}^n (n-r) \binom{n}{r}^2 = n \sum_{r=0}^n \binom{n}{r}^2 - \sum_{r=0}^n r \binom{n}{r}^2 \] \[ 2 \sum_{r=0}^n r \binom{n}{r}^2 = n \binom{2n}{n} \] \[ \sum_{r=0}^n r \binom{n}{r}^2 = \frac{n}{2} \binom{2n}{n} \]

Then

\[ S = \frac{4}{2n+3} \left( \frac{n}{2} \binom{2n}{n} \right) + \frac{3}{2n+3} \binom{2n}{n} = \frac{2n \binom{2n}{n} + 3 \binom{2n}{n}}{2n+3} = \frac{(2n+3)\binom{2n}{n}}{2n+3} = \binom{2n}{n} \]

Therefore,

\[ S = \binom{2n}{n} \]