Alg10.17

From Exampleproblems

Prove that ${C \choose 1}^{2}-2{C \choose 2}^{2}+3{C \choose 3}^{2}-.....+2n ({C \choose 2n})^2=(-1)^{n-1}(n\cdot 2n\!C_{n})\,$

$(1+x)^{2n}=1+2n\!C_1 x+2n\!C_2 x^2+......+2n\!C_{2n} x^{2n}\,$

Differentiating both sides w.r.t x

$2n(1+x)^{2n-1}=2n\!c_1+2n\!C_2 (2x)+......+2n\!C_{2n} (2nx^{2n-1})\,$

${C \choose 1}+{C \choose 2}(2x)+{C \choose 3}(3x^2)+.....+2n {C \choose 2n}x^{2n-1}\,$

${C \choose 1}+2{C \choose 2}x+3{C \choose 3} x^2+.....+2n {C \choose 2n} x^{2n-1}\,$

$(1-\frac{1}{x})^{2n}={C \choose 0}-{C \choose 1} \frac{1}{x}+{C \choose 2} \frac{1}{x^2}-.....+{C \choose 2n}\frac{1}{x^{2n}}\,$

Multiplying the above two equations,we get

$2n(1+x)^{2n-1} (1-\frac{1}{x})^{2n}=[{C \choose 1}+2{C \choose 2}x+.....+2n\!C_{2n} x^{2n-1}][{C \choose 0}-{C \choose 1} \frac{1}{x}+{C \choose 2}\frac{1}{x^2}-.....+{C \choose 2n} \frac{1}{x^{2n}}]\,$

In the above equation,the coefficient of $\frac{1}{x}\,$ in the RHS is

$-({C \choose 1}^{2})+2{C \choose 2}^{2}-3{C \choose 3}^{2}+....+2n\cdot ({C \choose 2n})^2\,$

Simplifying the LHS we get $\frac{2n(1-x^2)^{2n-1} (1-x)}{x^{2n}}\,$

Therefore,the coefficient of $\frac{1}{x}\,$ in the above expression is

is equal to the coefficient of $x^{2n-1} \,$ in $2n(1-x^2)^{2n-1} (1-x)\,$

which is equal to $2n(-1)^{n-1} (2n-1)\!C_{n-1} (-1)=(-1)^n\cdot (2n) \cdot \frac{(2n-1)!}{(n-1)!n!}=\frac{(-1)^n (2n)! n}{n(n-1)! n!}\,$ which is equal to

$(-1)^n\cdot n \frac{(2n)!}{(n!)^2}=(-1)^n\cdot n \cdot 2n\!C_n\,$

Therefore , $-({C \choose 1}^{2})+2{C \choose 2}^{2}-3{C \choose 3}^{2}+....+2n\cdot ({C \choose 2n})^2=(-1)^n\cdot n \frac{(2n)!}{(n!)^2}=(-1)^n\cdot n \cdot 2n\!C_n\,$

Main Page:Algebra

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