n-1-3-2-3-n-in-expanded-form-

Question Number 159775 by Ghaniy last updated on 22/Nov/21

\prod_{n = 1}^{\infty} \frac{α^{3} + β^{2}}{3^{n}} = ?

in expanded form

Answered by Canebulok last updated on 21/Nov/21

Solution :

(x^{3} + π^{2}) \cdot \underset{n = 1}{\prod^{100}} \frac{1}{4^{n}} = \underset{n = 1}{\prod^{100}} \frac{x^{3} + π^{2}}{4^{n}}

∴

(x^{3} + π^{2}) \cdot \underset{n = 1}{\prod^{100}} \frac{1}{4^{n}} = \frac{x^{3} + π^{2}}{4^{(\underset{n = 1}{\sum^{100}} n)}} = \frac{x^{3} + π^{2}}{4^{5050}}

Commented by Ghaniy last updated on 22/Nov/21

Thanks sir \dots . I {wasn}^{'} t sure

I thought the 100 will affect (α^{3} + β^{2}) as in \frac{{(α^{3} + β^{2})}^{\infty}}{3^{5050}}

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