Question 7 and 8, Exercise 4.8

Solutions of Question 7 and 8 of Exercise 4.8 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the sum of $n$ term of the series: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \times 10}+\ldots$$

Solution.

Given: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \times 10}+\dots$$ Let $T_k$ represents the kth term of the series. Then $$\begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*}$$ Resolving it into partial fraction: $$\begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*}$$ Multiplying with $(3k-2)(3k+1)$ $$\begin{align*} 1 = (3k+1)A+(3k-2)B \ldots (2) \end{align*}$$ Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have $$\begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*}$$ Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have $$\begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*}$$ Using value of $A$ and $B$ in (1), we get $$\begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{1}{3(3k-2)}-\frac{1}{3(3k+1)} \end{align*}$$ This gives $$\begin{align*} T_k &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*}$$ Taking sum $$\begin{align*} S_n&=\sum_{k=1}^n T_k =\frac{1}{3}\sum_{k=1}^n\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \\ & = \frac{1}{3}\left[\left(\frac{1}{1}-\frac{1}{4}\right) + \left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+\ldots\right.\\ &\left.+\ldots+\left(\frac{1}{3n-5}-\frac{1}{3n-2}\right)+\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right) \right] \\ & = \frac{1}{3}\left[1-\frac{1}{3n+1} \right] \\ & = \frac{1}{3}\left[\frac{3n+1-1}{3n+1} \right] \\ & = \frac{1}{3}\left[\frac{3n}{3n+1} \right] \\ & = \frac{n}{3n+1} \end{align*}$$ Hence the sum of given series is $\dfrac{n}{3n+1}$.

Find the sum of $n$ term of the series: $$\frac{1}{1 \times 6}+\frac{1}{6 \times 11}+\frac{1}{11 \times 16}+\ldots$$

Solution.

Let $T_k$ represents the kth term of the series. Then $$\begin{align*} T_k &=\frac{1}{(5k-4)(5k+1)}. \end{align*}$$ Resolving it into partial fraction: $$\begin{align*} \frac{1}{(5k-4)(5k+1)} = \frac{A}{5k-4}+\frac{B}{5k+1} \ldots (1) \end{align*}$$ Multiplying with $(5k-4)(5k+1)$ $$\begin{align*} 1 = (5k+1)A+(5k-4)B \ldots (2) \end{align*}$$ Put $5k-4=0$ $\implies k=\dfrac{4}{5}$ in (2), we have $$\begin{align*} &1 = \left(5\times\frac{4}{5}+1 \right)A+0 \\ \implies &A = \frac{1}{5}. \end{align*}$$ Now put $5k+1=0$ $\implies k=-\dfrac{1}{5}$ in (2), we have $$\begin{align*} &1 = 0+\left(5\left(-\frac{1}{5}\right)-4 \right)B \\ \implies &B = -\frac{1}{5}. \end{align*}$$ Using value of $A$ and $B$ in (1), we get $$\begin{align*} \frac{1}{(5k-4)(5k+1)} = \frac{1}{5(5k-4)}-\frac{1}{5(5k+1)} \end{align*}$$ This gives $$\begin{align*} T_k &=\frac{1}{5}\left[\frac{1}{5k-4}-\frac{1}{5k+1}\right] \end{align*}$$ Taking sum $$\begin{align*} S_n&=\sum_{k=1}^n T_k =\frac{1}{5}\sum_{k=1}^n\left[\frac{1}{5k-4}-\frac{1}{5k+1}\right] \\ & = \frac{1}{5}\left[\left(\frac{1}{1}-\frac{1}{6}\right) + \left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)\right.\\ &\left.+\ldots+\left(\frac{1}{5n-9}-\frac{1}{5n-4}\right)+\left(\frac{1}{5n-4}-\frac{1}{5n+1}\right) \right] \\ & = \frac{1}{5}\left[1-\frac{1}{5n+1} \right] \\ & = \frac{1}{5}\left[\frac{5n+1-1}{5n+1} \right] \\ & = \frac{1}{5}\left[\frac{5n}{5n+1} \right] \\ & = \frac{n}{5n+1}. \end{align*}$$ Hence the sum of given series is $\dfrac{n}{5n+1}$.

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