Question 3 and 4, Exercise 4.8

Solutions of Question 3 and 4 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.

Using the method of difference, find the sum of the series: $1+4+13+40+121+ \ldots$ to $n$ term.

Solution. Let $$S_{n}=1+4+13+40+121+\ldots +T_{n}$$ Also $$S_{n}=1+4+13+40+\ldots +T_{n-1}+T_{n}.$$ Subtracting the second expression from the first expression, we have $$\begin{align*} S_{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n} \\ & -\left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*}$$ $$\begin{align*} \implies 0=&1+(4-1)+(13-4)+(40-13)+(121-40) \\ & +\ldots+(T_{n}-T_{n-1})-T_{n}. \\ \implies 0=&1+(3+9+27+81+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*}$$ Then $$\begin{align*} T_{n} & =1+(3+9+27+81+\ldots \text { up to }(n-1) \text { terms }) \\ & =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =1+\frac{3^{n}-3}{2} \\ & =\frac{2+3^{n}-3}{2} \\ \\ & =\frac{3^{n}-1}{2}. \end{align*}$$ Thus, the kth term of the series: $$T_{k}=\frac{3^{k}-1}{2}.$$ Now taking the sum, we get $$\begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} \frac{3^{k}-1}{2} \\ & =\frac{1}{2}\left(\sum_{k=1}^{n} 3^{k} - \sum_{k=1}^{n} 1\right) \\ & =\frac{1}{2}\left((3+3^2+3^3+\ldots +3^{n})-n\right) \\ & =\frac{1}{2}\left(\frac{3(3^n-1)}{3-1}-n\right) \\ & =\frac{1}{2}\left(\frac{3^{n+1}-3}{2}-n\right) \\ & =\frac{1}{4}(3^{n+1}-3-2n). \end{align*}$$ Hence, the sum of the given series is $\frac{1}{4}(3^{n+1}-3-2n)$.

Using the method of difference, find the sum of the series: $1+2+4+7+11+16+\ldots$ to $n$ term.

Solution.

Let $$S_{n}=1+2+4+7+\ldots +T_{n}$$ Also $$S_{n}=1+2+4+7+\ldots +T_{n-1}+T_{n}.$$ Subtracting the second expression from the first expression, we have $$\begin{align*} S_{n}-S_{n}& =1+2+4+7+\ldots +T_{n} \\ & -\left(1+2+4+7+\ldots +T_{n-1}+T_{n}\right) \end{align*}$$ $$\begin{align*} \implies 0=&1+(2-1)+(4-2)+(7-4)+\ldots+(T_{n}-T_{n-1})-T_{n}. \\ \implies 0=&1+(1+2+3+ \ldots \text{ up to } (n-1) \text{ terms })-T_{n}. \end{align*}$$ Then $$\begin{align*} T_{n} & =1+(1+2+3+\ldots \text{ up to }(n-1) \text{ terms }) \\ & =1+\frac{n-1}{2}[2(1)+(n-1-1)(1)] \quad \left(\because S_n=\frac{n}{2}[2a + (n-1)d]\right) \\ & =1+\frac{(n-1)n}{2}. \\ & =\frac{2+n^2-n}{2}. \\ & =\frac{1}{2}(n^2-n+2). \end{align*}$$ Thus, the $k$-th term of the series is: $$T_{k}=\frac{1}{2}(k^2-k+2).$$ Now, taking the sum, we get $$\begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} \left(\frac{1}{2}(k^2-k+2)\right) \\ & = \frac{1}{2} \sum_{k=1}^{n} k^2 - \frac{1}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1\\ & =\frac{1}{2} \left(\frac{n(n+1)(2n+1)}{6}\right) - \frac{1}{2} \left(\frac{n(n+1)}{2}\right)+n \\ & =\frac{n}{12}\left((n+1)(2n+1)-3(n+1)+12 \right) \\ & =\frac{n}{12}\left(2n^2+2n+n+1-3n-3+12 \right) \\ & =\frac{n}{12}\left(2n^2+10 \right) \\ & =\frac{n}{6}\left(n^2+5 \right) \\ \end{align*}$$ Thus $$S_{n} = \frac{n}{6}\left(n^2+5 \right).$$

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