Let the n terms of series is 1-1-2 -3 -4-1-2-3 -4 -5-1-3-4 -5 -6- thenA- T n -1- n n -1 n -2 n -3B- S -1-6C- S -1-18D- S n -1-3 n -1 n -2 n -3

The correct options are A T_n =1n(n+1)(n+2)(n+3) C S_∞ =118General term of the series is T_n= 1n(n+1)(n+2)(n+3) T_n = 13 ( 1n(n+1)(n+2) 1(n+1)(n+2)(n+3) ) Let ...

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Standard XII

Mathematics

Method of Difference

Let the n ter...

Question

Let the $n$ terms of series is $\frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots \dots$ then

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

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S_{\infty} = \frac{1}{6}

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S_{\infty} = \frac{1}{18}

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

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Solution

The correct options are
A $T_{n} = \frac{1}{n (n + 1) (n + 2) (n + 3)}$
C $S_{\infty} = \frac{1}{18}$
General term of the series is
$T_{n} = \frac{1}{n (n + 1) (n + 2) (n + 3)}$
$T_{n} = \frac{1}{3} (\frac{1}{n (n + 1) (n + 2)} - \frac{1}{(n + 1) (n + 2) (n + 3)})$
Let $V_{n - 1} = \frac{1}{n (n + 1) (n + 2)}$
$∴ T_{n} = \frac{1}{3} [V_{n - 1} - V_{n}]$
Hence
$S_{n} = \frac{1}{3} [V_{0} - V_{n}] = \frac{1}{3} [\frac{1}{1.2.3} - \frac{1}{(n + 1) (n + 2) (n + 3)}]$
$S_{n} = \frac{1}{18} - \frac{1}{3 (n + 1) (n + 2) (n + 3)} S_{\infty} = \frac{1}{18}$

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Similar questions

Q. The sum of the series

1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

Q. Let the

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\frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots \dots

then

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n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

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Method of Difference

Standard XII Mathematics

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Let the n terms of series is 11.2.3.4+12.3.4.5+13.4.5.6+…… then

The correct options are A Tn=1n(n+1)(n+2)(n+3) C S∞=118General term of the series is Tn=1n(n+1)(n+2)(n+3) Tn=13(1n(n+1)(n+2)−1(n+1)(n+2)(n+3)) Let Vn−1=1n(n+1)(n+2) ∴Tn=13[Vn−1−Vn] Hence Sn=13[V0−Vn]=13[11.2.3−1(n+1)(n+2)(n+3)] Sn=118−13(n+1)(n+2)(n+3)S∞=118

Let the $n$ terms of series is $\frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots \dots$ then