1i guess we have to do it by telescoping summation....but getting no clue to break its nth term
9 Answers
1
341\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}= \frac{1}{2^n} +\frac{3n^2+10n+6}{2^n(n^4+4)} \right)
The first term is a GP to infinite series.
Let's look at the second term.
\frac{3n^2+10n+6}{2^n(n^4+4)} = 4 \left[ \frac{1}{2^n(n^2-2n+2)} - \frac{1}{2^{n+2}(n^2+2n+2)}\right]
=4 [f(n)-f(n+2)] where
f(n) = \frac{1}{2^n(n^2-2n+2)}
Now
\sum_{n=2}^{\infty} f(n) - f(n+2) = \sum_{n=2}^{\infty} [f(n)-f(n+1)]+ \sum_{n=2}^{\infty}[f(n+1) - f(n+2)]
Since \lim_{n \rightarrow \infty} f(n) = 0 the expression evaluates to
f(2)+f(3)=\frac{3}{20}
The entire expression then equals \frac{11}{10}
1ultimate solution sir !
may i ask what prompted the split in the 3rd line of thy solution
1if time permits also help here
the second one
http://www.targetiit.com/iit-jee-forum/posts/inequalities-18283.html
341The factorisation n^4+4 = (n^2-2n+2)(n^2+2n+2) is well known
Now to use telescopic summation we must obtain the general term as f(n+1)-f(n). That doesnt happen here. So in general we try to investigate that if f(n) = n2-2n+2, for what k do we have f(n+k) = n2+2n+2 and we easily see that k =2.
The rest is straightforward
For that inequality, I am trying it - I have reached a point where using Chebyshev inequality is the next logical step to finish the inequality, but I am not convinced of it. Let me look at it again
