Question 1 and 2, Exercise 4.8

Solutions of Question 1 and 2 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: $3+7+13+21+\ldots$ to $n$ term.

Solution.

Let $$ S_{n}=3+7+13+21+31+\ldots +T_{n} $$ Also $$ S_{n}=3+7+13+21+\ldots +T_{n-1}+T_{n}.$$ Subtracting second expression from the first expression, we have \begin{align*} S_{n}-S_{n}& =3+7+13+21+31+\ldots +T_{n} \\ & -\left(3+7+13+21+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(7-3)+(13-7)+(21-13) \\ &+(31-21)+\ldots +\left(T_{n}-T_{n-1}\right)-T_{n}. \\ \implies 0=&3+(4+6+8+10+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots \text { up to }(n-1) \text { terms }) \\ & =3+\frac{n-1}{2}[2(4)+(n-1-1)(2)] \quad\left(\because S_{n}=\frac{n}{2}[2 a+(n-1) d]\right) \\ & =3+\frac{n-1}{2}[8+(n-2)(2)] \\ & =3+\frac{n-1}{2}[2n+4] \\ & =3+(n-1)(n+2) \\ & =3+n^2-n+2n-2 \\ & =n^2+n+1 \end{align*} Thus, the kth term of series: $$ T_{k}=k^2++k+1 $$ Now taking sum, we get \begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} (k^2++k+1) \\ & =\sum_{k=1}^{n} k^2+\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1 \\ & =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n\\ & =\frac{n}{6}[(n+1)(2n+1)+3(n+1)+6] \\ & =\frac{n}{6}(2n^2+2n+n+1+3n+3+6) \\ & =\frac{n}{6}(2n^2+6n+10) \\ & =\frac{n}{3}(n^2+3n+5) \end{align*} Hence the sum of given series is $\dfrac{n}{3}(n^2+3n+5)$.

Using the method of difference, find the sum of the series: $1+4+10+22+\ldots$ to $n$ term.

Solution.

Let $$ S_{n}=1+4+10+22+46+\ldots +T_{n} $$ Also $$ S_{n}=1+4+10+22+\ldots +T_{n-1}+T_{n}. $$ Subtracting second expression from the first expression, we have \begin{align*} S_{n}-S_{n}& =1+4+10+22+46+\ldots +T_{n} \\ & -\left(1+4+10+22+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1)+(10-4)+(22-10)+(46-33) \\ & +\ldots+(T_{n}-T_{n-1})-T_{n}. \\ \implies 0=&1+(3+6+12+24+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+6+12+24+\ldots \text { up to }(n-1) \text { terms }) \\ & =1+\frac{3(2^{n-1}-1)}{2-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =1+3(2^{n-1}-1) \\ & =1+3\cdot 2^{n-1}-3 \\ & =3\cdot 2^{n-1}-2. \end{align*} Thus, the kth term of series: $$ T_{k}=3\cdot 2^{k-1}-2 $$ Now taking sum, we get \begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} (3\cdot 2^{k-1}-2) \\ & =3\sum_{k=1}^{n} 2^{k-1}- 2\sum_{k=1}^{n} 1 \\ & =3(1+2+2^2+2^3+... \text{ up to } n \text{ terms}) -2n\\ & =3\frac{(1)(2^n-1)}{2-1}-2n \\ & =3(2^n-1)-2n \end{align*} Hence the sum of given series is $3(2^n-1)-2n$.

Question 3 & 4 >