Alg10.11

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Prove that ({C \choose 0})^2+({C \choose 1})^2+({C \choose 2})^2+......+({C \choose n})^2=\frac{(2n)!}{(n!)^2}\,

(1+a)^n={C \choose 0}+{C \choose 1} a+{C \choose 2}a^2+{C \choose 3}a^3+......+{C \choose r}a^r+.......+{C \choose n} a^n\,

(a+1)^n={C \choose 0} a^n+{C \choose 1} a^{n-1}+{C \choose 2}a^{n-2}+.......+{C \choose r}a^{n-r}+.....+{C \choose n}\,

Therefore ((1+a)^n)((a+1)^n)=({C \choose 0}+{C \choose 1}a+{C \choose 2}a^2+{C \choose 3} a^3+......+{C \choose r} a^r+.......+{C \choose n} a^n)({C \choose 0} a^n+{C \choose 1} a^{n-1}+{C \choose 2} a^{n-2}+.......+{C \choose r} a^{n-r}+.....+C_n)\,

In the above, the coefficient of a^n\, in RHS is ({C \choose 0})^2+({C \choose 1})^2+({C \choose 2})^2+......+({C \choose r})^2+.......+({C \choose n})^2+.......\, Let this be 1.

But (1+a)^n (a+1)^n=(1+a)^{2n}\,, In this equation, the coeffcient of a^n\, is 2n\!C_n=\frac{(2n)!}{n!n!}=\frac{(2n!}{(n!)^2}\, let this be 2.

Comparing the coefficeints of a^n\, in 1 and 2. we get

({C \choose 0})^2+({C \choose 1})^2+({C \choose 2})^2+......+({C \choose r})^2+.......+({C \choose n})^2+.......=\frac{(2n!}{(n!)^2}\,


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