# Question 4 Review Exercise

Solutions of Question 4 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$

In the denominator Each term is the product of the successive terms of the sequence $1,4,7, \ldots$

Thus the general term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partial fractions \begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align} Multiplying both sides by $(3 n-2)(3 n+1)(3 n+4)$, we get \begin{align} 1=A(3 n+1)(3 n+4)+B(3 n-2)(3 n+4)+C(3 n-2)(3 n+1)& \\ =A[9 n^2+15 n+4]+B[9 n^2+6 n+8]+C[9 n^2-3 n-2]& \\ \Rightarrow[9 A+9 B+9 C] n^2+[15 A+6 B-3 C] n+[4 A+8 B-2 C]&=1\end{align} Comparing the coefficients of $n \cdot n$ and constant terms on the both sides of equations, we get $$A+B+C=0 \quad 15 A+6 B-3 C=0$$ $$4 A+8 B-2 C=1$$ Solving these three equations for the constants $A, B$ and $C$

we get $A=\dfrac{1}{18}, B=-\dfrac{1}{9}$ and $C=\dfrac{1}{18}$. Thus we have \begin{align} & a_n=\dfrac{1}{18(3 n-2)}-\dfrac{1}{9(3 n+1)}+\dfrac{1}{18(3 n+4)} \\ & a_n=\dfrac{1}{18(3 n-2)}-\dfrac{2}{18(3 n+1)}+\dfrac{1}{18(3 n+4)} \\ & a_n=\dfrac{1}{18}[\dfrac{1}{3 n-2}-\dfrac{2}{3 n+1}+\dfrac{1}{3 n+4}]\end{align} Taking summation of the both sides \begin{align}& \sum_{r=1}^n a_r \\ & =\dfrac{1}{18} \sum_{r=1}^n[\dfrac{1}{3 r-2}-\dfrac{2}{3 r+1}+\dfrac{1}{3 r+4}] \\ & =\dfrac{1}{18}[(1-\dfrac{2}{4}+\dfrac{1}{7})+(\dfrac{1}{4}-\dfrac{2}{7}+\dfrac{1}{10})+ \\ &\cdots+(\dfrac{1}{3 n-2}-\dfrac{2}{3 n+1}+\dfrac{1}{3 n+4})]\\ & =\dfrac{1}{18}[(1+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{10} \cdots+\dfrac{1}{3 n-2}) \\ & -(\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{2}{10}+\cdots+\dfrac{2}{3 n+1}) \\ & +(\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{13}+\cdots+\dfrac{1}{3 n+4})] \\ & =\dfrac{1}{18}[1+(\dfrac{1}{4}-\dfrac{2}{4})+(\dfrac{1}{7}-\dfrac{2}{7})+\cdots+ \\ & (\dfrac{1}{3 n-2}-\dfrac{2}{3 n-2})-\dfrac{2}{3 n+1} \\ & +\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{13}+\cdots+\dfrac{1}{3 n+4}] \\ & =\dfrac{1}{18}[1-\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}-\cdots-\dfrac{1}{3 n-2} \\ & .-\dfrac{2}{3 n+1}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{13}+\cdots+\dfrac{1}{3 n+4}] \\ & =\dfrac{1}{18}[1-\dfrac{1}{4}+(\dfrac{1}{7}-\dfrac{1}{7})+(\dfrac{1}{10}-\dfrac{1}{10})+ \\ & \cdots-\dfrac{1}{3 n+1}+\dfrac{1}{3 n+4}] \\ & =\dfrac{1}{18}[1-\dfrac{1}{4}-\dfrac{1}{3 n+1}+\dfrac{1}{3 n+4}]\end{align} Thus the sum of $n$ term is: $$S_n=\dfrac{1}{8}[1-\dfrac{1}{4}-\dfrac{1}{3 n+1}+\dfrac{1}{3 n+4}]$$ Taking limit $n \longrightarrow \infty$ we get $$S_{\infty}=\dfrac{1}{8}[1-\dfrac{1}{4}]=\dfrac{1}{24}$$