Summation this Summer! 3

Relevant wiki: Binomial Theorem

$\text{We will start with the following expansion :}$

$\large \begin{aligned} (1+x)^{2n}=\binom{2n}{0} + \binom{2n}{1}x+\binom{2n}{2}x^2+\cdots+\binom{2n}{2n}x^{2n}\end{aligned}$

$\text{Differentiating with respect to x we have }$

$\large \implies \begin{aligned} 2n(1+x)^{2n-1} = \binom{2n}{1}+2\binom{2n}{2}x+3\binom{2n}{3}x^2+\cdots+2n\binom{2n}{2n}x^{2n-1}\end{aligned}\text{ ........... (1)}$

$\large\begin{aligned} (x+1)^{2n} = \binom{2n}{0}x^{2n} + \binom{2n}{1}x^{2n-1}+\cdots+\binom{2n}{2n}\end{aligned}\text{ ........... (2)}$

Multiplying (1) & (2) we compare the co-efficient of $x^{2n-1}$

$\large 2n(1+x)^{4n-1} = (\color{#D61F06}{\binom{2n}{1}} + 2\color{#20A900}{\binom{2n}{2}x} + 3\color{#3D99F6}{\binom{2n}{3}x^2} + \cdots+2n\color{#69047E}{\binom{2n}{2n}x^{2n-1}})(\underbrace{\binom{2n}{0}x^{2n}}_{\color{#BA33D6}{\text{Rejected}}}+\color{#D61F06}{\binom{2n}{1}x^{2n-1}}+ \color{#20A900}{\binom{2n}{2}x^{2n-2}} + \color{#3D99F6}{\binom{2n}{3}x^{2n-3}} + \cdots+\color{#69047E}{\binom{2n}{2n}})$

The like colours in the parenthesis multiply and power of x in their product is $\text{2n-1}$ , therefore the co-efficient of $x^{2n-1}$ in RHS is our desired sum ,

$\large \mathfrak{S} = \dbinom{2n}{1}^2 +2.\dbinom{2n}{2}^2 +3.\dbinom{2n}{3}^2 \cdots \cdots + 2n.\dbinom{2n}{2n}^2$

$\text{Coefficient of } x^{2n-1} \text{ in } 2n(1+x)^{4n-1}\text{ is } 2n\binom{4n-1}{2n-1}$

The coefficients must be equal on both sides so,

$\large\mathfrak{S} = 2n\binom{4n-1}{2n-1} = 2n\frac{(4n-1)!}{(2n-1)!{2n}!}=\cancel{2n}\frac{(4n-1)!}{(2n-1)!\cancel{2n}(2n-1)!}=\boxed{\frac{(4n-1)!}{((2n-1)!)^2}}$

Let the sum be say equal to P.

P=0 (2nC0)^2+1 (2nC1)^2...........+2n-1 (2nC2n-1)^2+2n (2nC2n)^2

P=0 (2nC2n)^2+1 (2nC2n-1)^2...........+2n-1 (2nC1)^2+2n (2nC0)^2

add both the equations we get

2P=2n{(2nC0)^2+(2nC1)^2..........(2nC2n-1)^2+(2nC2n)^2}

P=n{4nC2n}

Hence P=(4n-1!)/(2n-1!)(2n-1!)

Just put n=1 to get the answer without solving it.