×

# NMTC Final test 2015

Hello friends, help me in finding the value of

$$\large\ a = 1 + \frac { 1 }{ { 2 }^{ 2 } } + \frac { 1 }{ { 3 }^{ 2 } } + ... + \frac { 1 }{ { 2015 }^{ 2 } }$$ .

Do post solution.

Note by Priyanshu Mishra
1 year, 6 months ago

Sort by:

Solution :

$$\displaystyle \sum^{\infty}_{n=1}\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6} \sim 1.64 \text{ and }\displaystyle \sum^{2015}_{n=1}\dfrac{1}{n^{2}}=a$$

$$\text{Simply by observation, we can conclude }\Rightarrow 1<a<1.64$$

$$\text{Therefore, }\lfloor a \rfloor=\boxed{1}$$ · 1 year, 6 months ago

Thank for a solution. But please explain me how you got this:

$$\large\ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } = \frac { \pi ^{ 2 } }{ 6 } }$$ ? · 1 year, 6 months ago

See this. · 1 year, 6 months ago

Thanks ,now i understood. However , how many points will you give for this problem? · 1 year, 6 months ago

75 of 400 · 1 year, 6 months ago

Well,it is obvious if one knows the solution of the basel problem .However this is a much simpler question.Notice that the first term is 1,the next two terms are less than 2×1÷2^2 that is 1/2,the next four terms less than 1/4 and so on.Even if we did this til infinity,the expression would be less than 1+1/2+1/4.... which is 2.Hence the required value is 1. · 1 year, 6 months ago

[a]=1 · 1 year, 6 months ago

Question has asked for value of a, not floor function of a. So a will be equal to approx. 1.64 . Do note that the series is going smaller and smaller. · 1 year, 5 months ago

In the question he didn't wrote about the integral value but in the original paper it was asked to find the integer part of $$a$$. · 1 year, 5 months ago

Yes, I forgot to write that.

By the way, how many questions you solved in the exam ? · 1 year, 5 months ago

I didn't gave the exam :-( · 1 year, 5 months ago

Ok, :) , integral value will be one only. · 1 year, 5 months ago