Question

In order to find the sum of $n$ terms of a series, each term of which is composed of $r$ factors in arithmetical progression or composed of the reciprocal of the product of $r$ factors in arithmetical progression, the first factor of the terms being the same, $A.P.$ for such type of series we have the following rules,
Rule A : (For .the sum $n$ terms of a series each terms of which is composed of as the product of $r$ factors which are in $A,P.$)
Write down $n^{th}$ term, affix the next factor at the end, divide by the number of factors thus increased and by common difference & then add a constant, Which is to be found by ascribing to $n$ some particular value
Rule : B (For the sum of $n$ terms of a series each term of which is composed of the reciprocal of the product of $r$ factors which are in $A.P$ ),
Write down the $n^{th}$ term, strike of a factor from the starting, divide by the number of factor so diminished & by the common difference, change the sign, then add a constant which is to be determined by ascribing to $n$ some particular value.
On the basis of above information answer the following questions,Let $S_n$ denotes the sum to $n$ terms of the series $\frac{1}{\mathrm{3.4.5}}+\frac{2}{\mathrm{4.5.6}}+\frac{3}{\mathrm{5.6.7}}+...$ then $\underset{x\to \infty }{lim}{S}_n$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{1}{20}$
• C. $\frac{1}{6}$
• D. $\frac{1}{60}$

Answer 1

The correct option is C $\frac{1}{6}$

GIven series Is $\frac{1}{\mathrm{3.4.5}}+\frac{2}{\mathrm{4.5.6}}+\frac{3}{\mathrm{5.6.7}}+...$
$\therefore t_n=\frac{n}{(n+2)(n+3)(n+4)}$ .......(A)
$\therefore t_n=\frac{1}{(n+3)(n+4)}-\frac{2}{(n+2)(n+3)(n+4)}$
$\therefore S_n=c-\frac{1}{n+4}+\frac{2}{2(n+3)(n+4)}$ ....(B)
Putting $n=1$
$\therefore S_1=t_1=\frac{1}{\mathrm{3.4.5}}=c-\frac{1}{5}+\frac{1}{20}$ from (A) & (B)
$\Rightarrow c=\frac{1}{6}$
$\therefore S_n=\frac{1}{6}-\frac{1}{n+4}+\frac{1}{(n+3)(n+4)}$
$\therefore \underset{n\to \infty }{lim}{S}_n=\frac{1}{6}$
Ans: C