Question

Evaluate the sum of n terms

$\frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + - - - + \frac{1}{n(n + 1)(n + 2)(n + 3)}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: \dfrac{1}{1.2.3.4} + \dfrac{1}{2.3.4.5} + - - - + \dfrac{1}{n(n + 1)(n + 2)(n + 3)}$

Let assume that $r^{th}$ term of the series be

$\rm \:T_{r} = \dfrac{1}{r(r + 1)(r + 2)(r + 3)}$

Now,

$\rm \:T_{r + 1} = \dfrac{1}{(r + 1)(r + 2)(r + 3)(r + 4)}$

can be rewritten as

$\rm \:T_{r + 1} = \dfrac{r}{r(r + 1)(r + 2)(r + 3)(r + 4)}$

$\rm \:T_{r + 1} = \dfrac{rT_{r}}{r + 4}$

$\rm \: (r + 4)T_{r + 1} = rT_{r}$

$\rm \: (r + 1 + 3)T_{r + 1} = rT_{r}$

$\rm \: (r + 1)T_{r + 1} + T_{r + 1}= rT_{r}$

$\rm \: rT_{r} - (r + 1)T_{r + 1} = 3T_{r + 1}$

On substituting r = 1, 2, 3, ..., n

$\rm \: T_{1} - 2T_{2} = 3T_{2}$

$\rm \: 2T_{2} - 3T_{3} = 3T_{3}$

$\rm \: 3T_{3} - 3T_{4} = 3T_{4}$

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$\rm \: nT_{n} - (n + 1)T_{n + 1} = 3T_{n + 1}$

On adding above all equations, we get

$\rm \: T_{1} - nT_{n} = 3(T_2 + T_3 + T_4 + - - - + T_n)$

can be further rewritten as

$\rm \:3T_1 + T_{1} - nT_{n} = 3(T_1 + T_2 + T_3 + T_4 + - - - + T_n)$

$\rm \:4T_1 - nT_{n} = 3(T_1 + T_2 + T_3 + T_4 + - - - + T_n)$

$\rm \:3(T_1 + T_2 + T_3 + - - + T_n)= 4 \times \dfrac{1}{1.2.3.4} - \dfrac{1}{n(n + 1)(n + 2)(n + 3)}$

$\rm \:3(T_1 + T_2 + T_3 + - - + T_n)= \dfrac{1}{6} - \dfrac{1}{n(n + 1)(n + 2)(n + 3)}$

$\rm \:T_1 + T_2 + T_3 + - - + T_n= \dfrac{1}{18} - \dfrac{1}{3n(n + 1)(n + 2)(n + 3)}$

Hence,

$\rm\implies \:S_n= \dfrac{1}{18} - \dfrac{1}{3n(n + 1)(n + 2)(n + 3)}$