The sum of the first n terms of the series 1-2-3-4-7-8-15-16- is

The correct option is C 1/2 + 3/4 + 7/8 + .... = ( 1 1/2 ) + ( 1 1/4 ) + ( 1 1/8 )+..... = n ( 1/2 + 3/4 + 7/8 + ...to n terms )a G.P = n 1/2 ( 1 1/2 ...

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Byju's Answer

Standard XII

Mathematics

Sigma n

The sum of th...

Question

The sum of the first n terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + . . .$ is

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Solution

The correct option is C
$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + . . . . = (1 - \frac{1}{2}) + (1 - \frac{1}{4}) + (1 - \frac{1}{8}) + . . . . . = n - (\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + . . . t o n t e r m s) a G . P = n - \frac{\frac{1}{2} (1 - \frac{1}{2^{n}})}{8} = n - 1 + 2^{- n}$

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The sum of the first n terms of the series 12+34+78+1516+... is

The correct option is C 12+34+78+....=(1−12)+(1−14)+(1−18)+.....=n−(12+34+78+...to n terms)a G.P=n−12(1−12n)8=n−1+2−n

The sum of the first n terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + . . .$ is

The correct option is C
$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + . . . . = (1 - \frac{1}{2}) + (1 - \frac{1}{4}) + (1 - \frac{1}{8}) + . . . . . = n - (\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + . . . t o n t e r m s) a G . P = n - \frac{\frac{1}{2} (1 - \frac{1}{2^{n}})}{8} = n - 1 + 2^{- n}$