Question

$\large \sf \frac{2 \sqrt{2} }{9801} \sum \limits_{n = 0}^{ \infty} \frac{(4n)!(1103 + 26390n)}{(n!) {}^{4} {396}^{4n} } = \frac{1}{\pi}$​

Answer 1

Step-by-step explanation:

given :

• 1/π = 2 √2 /9801 ∑∞k=0 (4 k ) (1103) + (26390 k) / ( k) 396

solution :

• please check the full attached file

Answer 2

Answer:

$\huge \dag \sf{Question}$

$\rm \red{\large \sf \frac{2 \sqrt{2} }{9801} \sum \limits_{n = 0}^{ \infty} \frac{(4n)!(1103 + 26390n)}{(n!) {}^{4} {396}^{4n} } = \frac{1}{\pi}}$

Answer :-

All answers →Refers← to the attachment

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