1sir dont u think that the third term should have 2^3 in the denominator.......
Find the value of
[\frac{n+1}{2}]+[\frac{n+2}{2^2}]+[\frac{n+4}{2^3}]+...+[\frac{n+2^k}{2^{k+1}}]+... sum goes to infinity...
Edit: [.] is the greatest integer Function
We have done something of this kind before..
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11 Answers
1
1if we write the kth term as : (n + 2k) / 2k+1
so,we can write its summation to get the answer : \sum_{k=0}^{infinity}{n / 2^k^+^1} + \sum_{k=0}^{infinity}{2^k / 2^k^+^1}
i think its rite unless i hv made som blunder
3Sandwich theorem will give us two inequalities but we can get our ans only if it gives us a tight inequality.
3also abhi u cant assume the nth term like as we dont know whether it is an integer,bcos at some term 'r' we have denominator gr8er than numerator so the rest term of the sequence will be zero.
1the general term is \sum_{0}^{inf }{(n+2^k)/2^(k+1)
=\sum_{0}^{inf }{[(n/2^(k+1))+1/2]
The first term will continue to be an integer until 2^(k+1)>n
or, k> log2 n - 1
so the rest of the terms r 0 since [.] is GIF.
this the conclusion i cud reach so far!!!
trying to work out more steps...
tell me if my approach is crrt!!!!
1S=\sum_{k=0}^{\inf}{\left[ \frac{n+2^{k}}{2^{k+1}}\right]} \\\texttt{From Hermites identity :} \\\left[ x\right]+\left[x+\frac{1}{2} \right]=[2x] \\S=\sum_{k=0}^{\inf}{\left[\frac{n}{2^k}\right]-\left[\frac{n}{2^{k+1}} \right]} \\\texttt{which telescopes to give :} \\{S=[n]}
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