lim-n-2-2-3-2-5-2-n-1-3-2-3-4-3-n-1-n-2-1-4-

Question Number 36969 by rahul 19 last updated on 07/Jun/18

{[\lim_{n \to \infty} {(2 . 2^{3} . 2^{5} \dots . . 2^{n - 1} . 3^{2} . 3^{4} \dots . . 3^{n})}^{\frac{1}{n^{2} + 1}}]}^{4} = ?

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

s = 1 + 3 + 5 + \dots + n - 1

n - 1 = 1 + (t - 1) 2

t = \frac{n - 2}{2} + 1

t = \frac{n}{2}

s = \frac{n}{4} {2 \times 1 + (\frac{n}{2} - 1) 2}

= \frac{n}{4} {2 + n - 2}

= \frac{n^{2}}{4}

s_{1} = 2 + 4 + \dots + n

n = 2 + (t_{1} - 1) 2

t_{1} = \frac{n}{2}

s_{1} = \frac{n}{4} {2 \times 2 + (\frac{n}{2} - 1) \times 2}

s_{1} = \frac{n}{4} {4 + n - 2}

s_{1} = \frac{n^{2} + 2 n}{4}

c o n t d

\lim_{n \to \infty} {2^{\frac{n^{2}}{4}} \times 3^{\frac{n^{2} + 2 n}{4}}}^{\frac{4}{n^{2} + 1}}

\lim_{n \to \infty} {2^{\frac{n^{2}}{n^{2} + 1}} \times 3^{\frac{n^{2} + 2 n}{n^{2} + 1}}}

\lim_{n \to \infty} {2^{\frac{1}{1 + \frac{1}{n^{2}}}} \times 3^{\frac{1 + \frac{2}{n}}{1 + \frac{1}{n^{2}}}}}

= 2 \times 3 = 6

Commented by rahul 19 last updated on 07/Jun/18

Thank you sir .

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

i w a n t t o s h a r e t h a t n e w t o n l a w s o r i g i n a t e d

i n i n d i a d o i s h a r e k n p u b l i c d o m a i n p l s r e p l y

Commented by rahul 19 last updated on 07/Jun/18

everything is welcome from your side sir! :)