1-1-2-1-2-3-1-3-4-1-nn-1 equals

The correct option is B Observe that this series is in the form of nbsp;1/a_1a_2 + 1/a_2a_3 nbsp;+.......+1/a_na_n+1, where a_1,a_2,.......are in A.P T_n ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

1/1.2+1/2.3+1...

Question

$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . . . + \frac{1}{n (n + 1)}$ equals

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Solution

The correct option is B

Observe that this series is in the form of $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . . . . . + \frac{1}{a_{n} a_{n + 1}}$ , where $a_{1}$ , $a_{2}$ ,.......are in A.P

$T_{n} = \frac{1}{(n t h t e r m o f 1, 2, . . . . .) (n t h t e r m o f 2, 3, . . . .)}$

= $\frac{1}{n (n + 1)}$

Let $V_{n}$ = $\frac{1}{(n + 1)}$ (leaving first factor Denominator of $T_{n}$ )

$V_{n - 1}$ = $\frac{1}{n}$

$V_{n}$ - $V_{n - 1}$ = $\frac{1}{n + 1}$ - $\frac{1}{n}$ = $\frac{- 1}{n (n + 1)}$ = - $T_{n}$ )

$S_{n} = \sum_{n = 1}^{n} T_{n} = - \sum_{n = 1}^{n} V_{n} - V_{n - 1}$

= - $[V_{n} - V_{0}]$ = 1 - $\frac{1}{1 + n}$ = $\frac{n}{n + 1}$

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Similar questions

lim x \to \infty (\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . + \frac{1}{n (n + 1)})

is equal to

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \dots \dots + \frac{1}{n (n + 1)} =

Solve:

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . + . \frac{1}{n (n + 1)}

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . + . . . . \frac{1}{n (n + 1)}

equals
[AMU 1995; RPET 1996; UPSEAT 1999, 2001]

Q. 1.2 + 2.3 + 3.4 + ... + n (n + 1) =

\frac{n (n + 1) (n + 2)}{3}

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11.2+12.3+13.4+.......+1n(n+1) equals

$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . . . + \frac{1}{n (n + 1)}$ equals