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Sum of Infinite Terms of a GP

Prove that :2...

Question

Prove that :

$(2^{\frac{1}{4}} {.4}^{\frac{2}{8}} {.8}^{\frac{3}{16}} {.16}^{\frac{4}{32}} . . . . \infty) = 2.$

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Solution

LHS = $2^{\frac{1}{4}} {.4}^{\frac{1}{8}} {.8}^{\frac{1}{16}} {.16}^{\frac{1}{32}} . . . . \infty = 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} . \infty)} = 2^{(\frac{1}{2^{2}} + \frac{2}{2^{3}} + \frac{3}{2^{4}} + \frac{4}{2^{5}} + . . . . \infty)} = 2^{\frac{1}{2^{2}} {1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{2^{3}} . . . \infty}} = 2^{\frac{1}{2^{2}} ⎧ ⎪ ⎨ ⎪ ⎩ \frac{1}{1 - \frac{1}{2}} + \frac{1. \frac{1}{2}}{{(1 - \frac{1}{2})}^{2}} ⎫ ⎪ ⎬ ⎪ ⎭} = 2^{2^{\frac{1}{2} {2 + 2}}} = 2^{1} = 2$ = RHS

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Standard XII Mathematics