The sum of the series 1-3- 7-1-7- 11-1-11- 15- is

Explanation for the correct optionLet, S=1/3× 7+1/7× 11+1/11× 15+........⇒ S=1/4×(1/3 1/7)+1/4(1/7 1/11)+1/4(1/11 1/15)+........0ex0ex⇒ S=1/4(1/3 1/7+1/7 1/11+1 ...

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Byju's Answer

Standard VIII

Mathematics

Commutative Property of Rational Numbers

The sum of th...

Question

The sum of the series $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + . . . . . . . .$ is

$\frac{1}{3}$

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$\frac{1}{6}$

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$\frac{1}{9}$

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$\frac{1}{12}$

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Solution

The correct option is D
$\frac{1}{12}$

Explanation for the correct option
Let, $S = \frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + . . . . . . . .$
$\Rightarrow S = \frac{1}{4} \times (\frac{1}{3} - \frac{1}{7}) + \frac{1}{4} (\frac{1}{7} - \frac{1}{11}) + \frac{1}{4} (\frac{1}{11} - \frac{1}{15}) + . . . . . . . . \Rightarrow S = \frac{1}{4} (\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \frac{1}{11} - \frac{1}{15} + . . . . . . . . .) \Rightarrow S = \frac{1}{4} (\frac{1}{3}) \Rightarrow S = \frac{1}{12}$
Hence, $Option (D)$ is correct.

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Similar questions

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. If

ω

is complex cube root of unity and a, b, c are three real numbers such that

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

and

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω

then

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

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The sum of the series 13×7+17×11+111×15+........ is

The correct option is D 112Explanation for the correct optionLet, S=13×7+17×11+111×15+........⇒S=14×13-17+1417-111+14111-115+........⇒S=1413-17+17-111+111-115+.........⇒S=1413⇒S=112Hence, OptionD is correct.

The sum of the series $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + . . . . . . . .$ is