The sum of the series 3-4-8-3-5-4-8-12-3-5-7-4-8-12-16- is

Explanation for the correct option Given: 3/4·8 3·5/4·8·12+3·5·7/4·8·12·16 ..........Let, S=3/4·8 3·5/4·8·12+3·5·7/4·8·12·16 ..........add 1 1/4to both sides⇒ S ...

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Byju's Answer

Standard VII

Mathematics

Addition of Rational Numbers

The sum of th...

Question

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . . . . . .$ is

$\sqrt{\frac{3}{2}} - \frac{3}{4}$

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$\sqrt{\frac{2}{3}} - \frac{3}{4}$

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$\sqrt{\frac{3}{2}} - \frac{1}{4}$

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$\sqrt{\frac{2}{3}} - \frac{1}{4}$

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Solution

The correct option is B
$\sqrt{\frac{2}{3}} - \frac{3}{4}$

Explanation for the correct option
Given: $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . . . . . .$
Let, $S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . . . . . .$
add $1 - \frac{1}{4}$ to both sides
$\Rightarrow S + 1 - \frac{1}{4} = 1 - \frac{1}{4} + \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . . . . . . \Rightarrow S + \frac{3}{4} = 1 - (\frac{1}{1}) \cdot (\frac{1}{4}) + (\frac{1}{4}) (\frac{3}{8}) - (\frac{1}{4}) (\frac{3}{8}) (\frac{5}{12}) + (\frac{1}{4}) (\frac{3}{8}) (\frac{5}{12}) (\frac{7}{16}) - . . . . . . . . \Rightarrow S + \frac{3}{4} = 1 - (\frac{1}{1}) \cdot (\frac{1}{4}) + (\frac{1 \cdot 3}{1 \cdot 2}) {(\frac{1}{4})}^{2} - (\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3}) {(\frac{1}{4})}^{3} + . . . . . . . . . . . . . . . .$
Let $p = 1, q = 2, x = \frac{1}{2}$
$S + \frac{3}{4} = 1 - (\frac{p}{1}) (\frac{x}{q}) + (\frac{p \cdot (p + q)}{1 \cdot 2}) {(\frac{x}{q})}^{2} - (\frac{p \cdot (p + q) (p + 2 q)}{1 \cdot 2 \cdot 3}) {(\frac{x}{q})}^{3} + . . . . . . . . . . . .$
expansion of ${(1 + x)}^{- \frac{p}{q}} = 1 - (\frac{p}{1}) (\frac{x}{q}) + (\frac{p \cdot (p + q)}{1 \cdot 2}) {(\frac{x}{q})}^{2} - (\frac{p \cdot (p + q) (p + 2 q)}{1 \cdot 2 \cdot 3}) {(\frac{x}{q})}^{3} + . . . . . . . . . . . .$
$S + \frac{3}{4} = {(1 + x)}^{- \frac{p}{q}}$
Put $p = 1, q = 2, x = \frac{1}{2}$
$\Rightarrow S + \frac{3}{4} = {(1 + \frac{1}{2})}^{- \frac{1}{2}} \Rightarrow S + \frac{3}{4} = \sqrt{{(\frac{3}{2})}^{- 1}} \Rightarrow S + \frac{3}{4} = \sqrt{\frac{2}{3}} \Rightarrow S = \sqrt{\frac{2}{3}} - \frac{3}{4}$
Hence, $Option (B)$ is correct.

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Similar questions

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) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. Mark whether true or false.
(a) 8 (7 + 3) = 8 × 7 + 8 × 3 ____
(b) 12 (16 + 4) = 12 × 16 + 12 × 4 ____
(c) 14 (5 + 2) = 14 × 5 + 14 × 2 ____
(d) 9 (8 − 3) = 9 × 8 − 9 × 3 ____

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

Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

Fill in the blanks:

$(i) - 4 \times \frac{7}{9} = \frac{7}{9} \times$ ___

$(i i) \frac{5}{11} \times \frac{- 3}{8} = \frac{- 3}{8} \times$ ___

$(i i i) \frac{1}{2} \times (\frac{3}{4} + \frac{- 5}{12}) = \frac{1}{2} \times$ ____ $+ (\times) \frac{- 5}{12}$

$(i v) \frac{- 4}{5} \times \frac{5}{7} \times$ ___ $= (\frac{- 4}{5} \times \frac{5}{7}) \times \frac{- 8}{9}$

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The sum of the series 34·8-3·54·8·12+3·5·74·8·12·16-.......... is

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - . . . . . . . . . .$ is