Question

please solve this question ☺️☺️☺️
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Answer 1

$\star\:\:\:\sf\large\underline\blue{Question:-}$

• Prove that $\sf {2}^{ \frac{1}{4} } \times {4}^{ \frac{1}{8}} \times {8}^{ \frac{1}{16} } \times ..... \infty = 2$

$\star\:\:\:\sf\large\underline\blue{Proof:-}$

$\sf {2}^{ \frac{1}{4} } \times {4}^{ \frac{1}{8}} \times {8}^{ \frac{1}{16} } \times ..... \infty$

$\sf= {2}^{ \frac{1}{4} } \times { ({2}^{2} )}^{ \frac{1}{8} } \times { ({2}^{3} )}^{ \frac{1}{16} } \times .... \infty$

$\sf = {2}^{ \frac{1}{4} } \times {2}^{ \frac{2}{8} } \times {2}^{ \frac{3}{16} } \times ... \infty$

$\sf = {2}^{ \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ... \infty }$

$\sf Let \: S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ... \infty$

Therefore,

$\sf \frac{1}{2} S = \frac{1}{8} + \frac{2}{16} + ... \infty$

$\sf \implies S - \frac{1}{2} S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + .... \infty$

$\sf \implies \frac{1}{2} S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + .... \infty$

$\sf \implies S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... \infty$

$\sf \implies 2S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... \infty$

$\sf \implies2S - S = 1$

$\sf \implies S = 1$

Therefore,

$\sf {2}^{S} = {2}^{1} = 2$

So,

$\boxed{ \sf {2}^{ \frac{1}{4} } \times {4}^{ \frac{1}{8}} \times {8}^{ \frac{1}{16} } \times ..... \infty = 2}$

Hence Proved.

Have a great day ahead.

Answer 2

Answer:

Reffered to the attachment

Step-by-step explanation:

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Warm regards:Miss Chikchiki