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- Question 1 Exercise 5.3
- ===Question 1(i)==== Find the sum of the series $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractio
- Question 4 Review Exercise
- Pakistan. =====Question 4===== Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product of ... $ Thus the general term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partia
- Question 1 Exercise 5.2
- & =(2+2^2+2^3++\ldots+2^n)-n \cdot 2^{n+1} \\ & =\dfrac{2(2^n-1)}{2-1}-n \cdot 2^{n+1} \\ & =2^{n+1}-2-n ... x^2+3 x^3+\ldots +3 x^{n-1}-(3 n-2) x^n \\ & =1+\dfrac{3 x(1-x^{n-1})}{1-x}-(3 n-2) x^n \\ S_n&=\dfrac{1-(3 n-2) x^n}{1-x}+\dfrac{3 x(1-x^{n-1})}{(1-x)^2}\end{align} =====Question 1(iii)===== Sum up to $n$
- Question 2 & 3 Exercise 5.4
- ion 2===== Find sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2} \\ S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+6 k-3 k-2} \\ & =\sum_{k=1}^n \dfrac{1}{3 k(3 k+2)-1(3 k+2)} \\ S_n&=\sum_{k=1}^n \dfrac{1}{(3
- Question 8 Review Exercise
- ^n a_r&=\sum_{r=1}^n r^3+\sum_{r=1}^n 3^r \\ & =[\dfrac{n(n+1)}{2}]^2+\dfrac{3(3^n-1)}{3-1} \\ & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align} Thus the sum of $n$ terms is: $$S_n=\dfrac{n^2(n+1)^2}{4}
- Question 4 & 5 Exercise 5.2
- f the following arithtical geometrical series $5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots$ ====Solution==== Let \begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(i) \
- Question 1 Exercise 5.1
- 1}^n j^2-4 \sum_{j=1}^n j+\sum_{j=1}^n 1 \\ & =4 \dfrac{n(n+1)(2 n+1)}{6}-4 \dfrac{n(n+1)}{2}+n \\ & =n[\dfrac{4(n+1)(2 n+1)-12(n+1)+6}{6}] \\ & =n[\dfrac{4(2 n^2+3 n+1)-12 n-12+6}{6}] \\ & =n[\dfrac{8 n^2+12n+4-12
- Question 5 & 6 Review Exercise
- ]-(7 n-1) x^n \\ \Rightarrow(1-x) S_n&=5+7 \cdot \dfrac{x(1-x^{n-1})}{1-x}-(7 n-1) x^n \\ \Rightarrow S_n&=\dfrac{5}{1-x}+\dfrac{7(x-x^n)}{(1-x)^2}-\dfrac{(7 n-1) x^n}{1-x}\\ \end{align} =====Question 6===== Sum the series: $\dfrac{
- Question 9 Review Exercise
- et \begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & =\dfrac{n-1}{2}[2 \cdot 4+2 \cdot(n-2)] \\ & =\dfrac{n-1}{2}[8+2 n-4] \\ & =\dfrac{(n-1)(2 n+4)}{2} \\ & =(n-1)(n+2) \\ \Rightarrow a_n&=n^2+n-2+a_1 \\ \Righ... _{r=1}^n r^2+\sum_{r=1}^n r+\sum_{r=1}^n 1 \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2}+n \\ & =n \cd
- Question 10 Review Exercise
- erm and the sum to $n$ terms of the series $1+(1+\dfrac{1}{2})+(1+\dfrac{1}{2}+\dfrac{1}{4})+(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8})+\ldots$ ====Solution==== The general term of the given
- Question 4 Exercise 5.4
- 4===== Find the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$ ====Solution==== Let \begin{align}S_n &=\sum_{k=1}^n \dfrac{1}{k^2+7 k+12} \\ & =\sum_{k=1}^n \dfrac{1}{(k+3)(k+4)}\end{align} Consider the $n^{\text {th }}$ term of the series $$u_n=\dfrac{1}{(n+3)(n+4)}$$ Resolving into partial fractions
- Question 2 & 3 Review Exercise
- =1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2} \cdot \dfrac{2 n+1+3}{3} \\ & =\dfrac{n(n+1)}{2}
- Question 7 Review Exercise
- 1}^n r^3+\sum_{r=1}^n r^2-\sum_{r=1}^n 1 \\ & =2[\dfrac{n(n+1)}{2}]^2+\dfrac{n(n+1)(2 n+1)}{6}-n \\ & =\dfrac{n^2(n+1)^2}{2}+\dfrac{n(n+1)(2 n+1)}{6}-n \\ & =\dfrac{3 n^2(n+1)^2}{6}+\dfrac{n(n+1)(2 n+1)}{6}-n \\ &
- Question 4 & 5 Exercise 5.1
- eral term of the sequence is: \begin{align}& T_j=\dfrac{j}{2}[2(2)+3(j-1)]\\ &=\dfrac{j(3 j+1)}{2} \\ & =\dfrac{1}{2}(3 j^2+j)\end{align} Taking sum of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_i=\dfrac{1}{2}[3 \sum_{j=1}^n j^2+\sum_{j=1}^n j] \\ & =\d
- Question 6 Exercise 5.1
- n j^3+3 \sum_{j=1}^n j^2+2 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6}+2 \dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{n(n+1)}{2}+3 \dfrac{(2 n+1)}{3}+2] \\ & =\dfrac{n(n+1)}{2}[\df