Sum 1 - 34 - 716 - 1564 - 31256 - - to infinity-

The correct option is C 83 Let S = 1+34+716+1564+31256 #160;...... ∞ Multiply both sides by 14 , we get S14 = 0+14+316+764+15256 ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

Sum 1 + 34 ...

Question

Sum $1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + . . . .$ to infinity.

\frac{4}{3}

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\frac{5}{3}

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\frac{8}{3}

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\frac{5}{4}

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Solution

The correct option is C $\frac{8}{3}$
Let $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256}$ ...... $\infty$
Multiply both sides by $\frac{1}{4}$ , we get
$S \frac{1}{4} = 0 + \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256}$ ........ $\infty$
Subtracting both equations, we get
$S \frac{3}{4} = 1 + \frac{2}{4} + \frac{4}{16} + \frac{8}{64} + \frac{16}{256}$ $\infty$
$\Rightarrow S \frac{3}{4} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$ ....... $\infty$
Clearly above series is G.P
Common ratio $=$ $\frac{1}{2}$
$\Rightarrow S \frac{3}{4} = \frac{1}{1 - \frac{1}{2}} = 2$
$\Rightarrow S = \frac{8}{3}$

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Similar questions

Q. Sum the following series

1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + . . . .

to infinity.

Q. Sum

1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + . . . . .

to infinity.

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

Q. The sum of the series

\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . .

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Sum 1+34+716+1564+31256+.... to infinity.

The correct option is C 83Let S=1+34+716+1564+31256 ......∞Multiply both sides by 14, we getS14=0+14+316+764+15256 ........∞Subtracting both equations, we getS34=1+24+416+864+16256 ∞⇒S34=1+12+14+18+116 .......∞Clearly above series is G.PCommon ratio = 12⇒S34=11−12=2⇒S=83

Sum $1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + . . . .$ to infinity.