Let \[ \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^{4}+1}}-\frac{2 n}{\left(n^{2}+1\right) \sqrt{n^{4}+1}}+\frac{n}{\sqrt{n^{4}+16}}-\frac{8 n}{\left(n^{2}+4\right) \sqrt{n^{4}+16}} +\ldots \ldots .+\frac{n}{\sqrt{n^{4}+n^{4}}}-\frac{2 n \cdot n^{2}}{\left(n^{2}+n^{2}\right) \sqrt{n^{4}+n^{4}}}\right) \text { be } \frac{\pi}{k}, \] using only the principal values of the inverse trigonometric functions. Then \(\mathrm{k}^{2}\) is equal to ______.

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Limits

Let

n \to \infty lim (\frac{n}{n ^{4} + 1} - \frac{2 n}{( n ^{2} + 1 ) n ^{4} + 1} + \frac{n}{n ^{4} + 16} - \frac{8 n}{( n ^{2} + 4 ) n ^{4} + 16} + \dots\dots . + \frac{n}{n ^{4} + n ^{4}} - \frac{2 n \cdot n ^{2}}{( n ^{2} + n ^{2} ) n ^{4} + n ^{4}}) be \frac{π}{k},

using only the principal values of the inverse trigonometric functions. Then

k^{2}

is equal to ______.

Answer: 32

The

r

-th term, divided through by

n^{2}

, is

\frac{1}{1 + ( r / n ) ^{4}} - \frac{2 ( r / n ) ^{2}}{( 1 + ( r / n ) ^{2} ) 1 + ( r / n ) ^{4}}

, so the sum

\to \int_{0}^{1} \frac{1 - x ^{2}}{( 1 + x ^{2} ) 1 + x ^{4}} d x

. With

t = x + \frac{1}{x}

(so

t^{2} - 2 = x^{2} + \frac{1}{x ^{2}}

) this is

\int_{2}^{\infty} \frac{d t}{t t ^{2} - 2} = \frac{1}{2} (\frac{π}{2} - \frac{π}{4}) = \frac{π}{4 2}

. Hence

k = 42

and

k^{2} = 32

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Integer answer