# Doubt from Trigonometry and Series

for first option c

Yes,how?

graph for n=1 nd n=2 u will se result

Done that already,I'd like a general solution

Note that when n^2-n+1 \le m \le n^2+n , the integer closest to \sqrt m is n

as n^2-n+1 = (n-1)^2+\dfrac{3}{4} so that n\ge \sqrt{n^2-n+1} >n-\dfrac{1}{2}

So given summation may be written as

\displaystyle \sum_{n=1}^{\infty} (2^n+2^{-n}) \left[\dfrac{1}{2^{n^2-n+1}}+\dfrac{1}{2^{n^2-n+2}} +\cdots +\dfrac{1}{2^{n^2+n}}\right] which using GP formula and simplifying becomes

2 \displaystyle \sum_{n=1}^{\infty}\dfrac{(2^{2n}+1)(2^{2n}-1)}{2^{(n+1)^2}} = 2\sum_{n=1}^{\infty}\dfrac{2^{4n}-1}{2^{(n+1)^2}} = 2\sum_{n=1}^{\infty}\dfrac{1}{2^{{(n-1)}^2}} - \dfrac{1}{2^{{(n+1)}^2}}

= \displaystyle 2\sum_{n=1}^{\infty} \left(\dfrac{1}{2^{{(n-1)}^2}} -\dfrac{1}{2^{{n}^2}} \right)+ \left(\dfrac{1}{2^{{n}^2}} -\dfrac{1}{2^{{(n+1)}^2}} \right)
which is the sum of two telescoping sums where the first converges to 1 and the second to \dfrac{1}{2} so that the sum is 2 \times \dfrac{3}{2}=3

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f(x) = a_0+a_1 \cos x+a_2 \cos 2x+\cdots +a_n \cos n x

We will assume that a_0>0

Let x_k= \dfrac{k \pi}{n}

Then \displaystyle f(x_{2k})= a_0+a_1 \cos x_{2k} +a_2 \cos2 x_{2k}+ \cdots + a_n

and since a_k \cos \theta > -|a_k| we have that the above quantity is

> a_n - (|a_0|+|a_1|+\ldots +|a_n| )>0

Again f(x_{2k+1})= a_0+a_1 \cos x_{2k+1} +a_2 \cos2 x_{2k+1}+ \cdots- a_n

< (a_0-|a_0|)+ (a_1 \cos x_{2k+1} - |a_1|+\ldots <0

Hence by Intermediate Value Theorem we always have at least one root between x_{2k} and x_{2k+1} i.e. between \dfrac{2 k \pi}{n} and \dfrac{(2 k+1) \pi}{n}

Thus every interval \left (\dfrac{m \pi}{n}, \dfrac{(m+1) \pi}{n} \right) has atleast one root. In the given interval we have 0 \le m \le 2n-1 and thus we have at least 2n roots

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Typo in second line n^2-n+1 = \left(n-\dfrac{1}{2}\right)^2+\dfrac{3}{4}

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sir my try little shorter