Find till infinity: 1 - n^2 + {n^2(n^2-1^2)}/(2!)^2 - {n^2(n^2-1^2)(n^2-2^2)}/(3!)^2................................ to infinity,where n belongs to N.
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1357The question is:
1-n^{2}+\frac{n^{2}(n^{2}-1^{2})}{2!^{2}}-\frac{n^{2}(n^{2}-1^{2})(n^{2}-2^{2})}{3!^{2}}....where n belongs to N
Sushovan Halder yes sir.Upvote·0· Reply ·2013-11-07 06:39:49
2305General term of this sum =
\frac{n^2(n^2-1^2)(n^2-2^2)\cdots (n^2-r^2)}{(r+1)!}
=\frac{(n+r)(n+r-1)\cdots (n+1)(n)(n)(n-1)\cdots (n-r+1)(n-r)}{(r+1)!^2}
=\binom{n+r}{r+1}\binom{n}{r+1}
(1-x^{-1})^{-n}=1+nx^{-1}+\binom{n+1}{2}x^{-2}+\cdots
(1-x)^{n}=1-nx+\binom{n}{2}x^2-\cdots
Therefore the constant term in
(1-x^{-1})^{-n}\cdot (1-x)^{n} \text{ is } =1-n^2+\frac{n^2(n^2-1^2)}{2}-\cdots
\text{But }(1-x^{-1})^{-n}\cdot (1-x)^{n}=(-x)^n\text{ has no constant term.}
\therefore 1-n^2+\frac{n^2(n^2-1^2)}{2}-\cdots =0
- Sushovan Halder thanks