1 Answer
Solution :-
The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x \( \dfrac{1}{2} \)
= 64[4 + 2 + 1 + \( \dfrac{1}{2} \)+ \( \dfrac{1}{2^2} \)]
Now, it’s seen that
4, 2, 1, \( \dfrac{1}{2} \), \( \dfrac{1}{2^2} \) is a G.P.
With first term, a = 4
Common ratio, r =\( \dfrac{1}{2} \)
We know,
\( s_n=\dfrac{a(1-r^n)}{1-r}\)
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= \(8( \dfrac{32-1}{32})=\dfrac{31}{4}\)
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