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- Question 1 Exercise 5.2
- +3.2^3+4.2^4+\ldots$. ====Solution==== Let \begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4+\ldots +... 2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(4-3) 2^3+... cdot 2^{n+1}-2^{n+1} \\ S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ terms th
- Question 1 Exercise 5.3
- $A+B=0 \text{and} A=1$$ Putting $A=1$,then \begin{align}1+B&=0\\ B&=-1\end{align} hence \begin{align} & \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} \\ & \Rightarrow T_n=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking summation of the both sides of the above
- Question 1 Exercise 5.1
- e both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2 \\ & =\s... frac{8 n^2-2}{6}] \\ & =\dfrac{n(4 n^2-1)}{3}\end{align} =====Question 1(ii)===== Sum the series $1^2+(1... 2+2^2$, $T_3=1^2+2^2+3^2$ and so on we get \begin{align}& T_j=1^2+2^2+3^2+\ldots+j^2 \\ & =\dfrac{j(j+1)(2 j+1)}{6}\end{align} Taking of sum of the both sides of the above, we
- Question 8 Review Exercise
- 3+3^n$$ Taking summation of the both sides \begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_{r=1}^n 3... & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align} Thus the sum of $n$ terms is: $$S_n=\dfrac{n^2(n... 2+3 n$$ Taking summation on the both sides \begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^2+3 \sum_{r=1... c{2(2n+1)+9}{6}\\ & =\dfrac{n(n+1)(4n+11)}{6}\end{align} Thus sum to $n$ terms is: $$S_n=\dfrac{n(n+1)(4n
- Question 9 Review Exercise
- es to compute the sum of the given series. \begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=6 \\ & a_4-a_3... xt { term of the series } \\ & 4,6,8, \ldots \end{align} Adding column wise, we get \begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & =\dfrac{n-1}{2}[2 \cdot 4+2 \cd... +3 \because a_1=3 \\ \Rightarrow a_n&=n^2+n+1\end{align} Taking summation of the both sides \begin{align}
- Question 4 & 5 Exercise 5.2
- frac{11}{3^3}+\ldots$ ====Solution==== Let \begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfra... }+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2}{3}+\dfrac{2}{... \ & =5+2 \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}\end{align} $\because$ the series is geometric $r=\dfrac{1}{
- Question 7 & 8 Exercise 5.1
- l term of the series is: $T_j=j(j+4)(j+8)$ \begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \sum_{j=1... +9)+8(n+9)] \\ & =\dfrac{n(n+1)(n+8)(n+9)}{4}\end{align} =====Question 8===== Find the sum of the series
- Question 2 & 3 Exercise 5.2
- 7^2 x^3+\ldots, x<1$. ====Solution==== Let \begin{align} & S_{\infty}=1+3^2 x+5^2 x^2+7^2 x^3+\ldots ..(1... \infty}=x+3^2 x^2+5^2 x^3+7^2 x^4+\ldots..(2)\end{align} Subtracting the (2) from (2), we get \begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2) x+(5^2-3^2) x^2+(7^2... x) S_{\infty}=x+8 x^2+16 x^3+24 x^4+\ldots(4)\end{align} Again subtracting (4) from (3) \begin{align} & [
- Question 2 & 3 Exercise 5.4
- n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+3... S_n&=\sum_{k=1}^n \dfrac{1}{(3 k-1)(3 k-2)} \end{align} The $n$ term of the above series is: $$u_n=\dfra... frac{B}{3 k+2}$$ Multiplying both sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & 1=A(3 k+2... +B(3 k-1) \\ & \Rightarrow(3 A+3 B) k+2 A-B=1\end{align} Comparing the cocfficients of $k$ and constants
- Question 4 Review Exercise
- 3 n+4)}$$ 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... 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 co
- Question 5 & 6 Review Exercise
- \ldots$ to $n$ terms. ====Solution==== Let \begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}.... ^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align} Subtracting the (ii) from (i) we get \begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\ &+[7 n-2-(... (x-x^n)}{(1-x)^2}-\dfrac{(7 n-1) x^n}{1-x}\\ \end{align} =====Question 6===== Sum the series: $\dfrac{1}
- Question 4 & 5 Exercise 5.1
- n==== The general 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^... c{n(n+1)}{4}[2 n+2] \\ & =\dfrac{n^2}{2}(n+1)\end{align} =====Question 5===== Sum: $2+5+10+17+\ldots$ to
- Question 2 Exercise 5.3
- 4+14+30+52+80+114+\ldots$ ====Solution==== \begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16 \\ & a_4... \text{ term of the sequence} 10,16,22, \ldots\end{align} which is a A.P. Adding column wise, we get \begin{align} a_n-a_1&=10+16+22+\ldots+(n-1) \text { terms } \... ad \because a_1=4 \\ \Rightarrow a_n&=3 n^2+n\end{align} Taking summation of the both sides \begin{align}
- Question 3 Exercise 5.3
- on==== We use the method of difference as: \begin{align} & a_2-a_1=10-4=6 \\ & a_3-a_2=18-10=8 \\ & a_4-a... \mathrm{n}-1) \text { term of the sequence } \end{align} $6,10,8, \ldots$ which is a A.P. Adding column wise, we get \begin{align}& a_n-a_1=6+10+8-\ldots +(n-1) \text { terms } \\... d \because a_1=4 \\ & \Rightarrow a_n=n^2+3 n\end{align} Taking summation of the both sides \begin{align}
- Question 4 Exercise 5.3
- $3+5+11+29+83+245+\ldots$ ====Solution==== \begin{align} & a_2-a_1=5-3=2 \\ & a_3-a_2=11-5=6 \\ & a_4-a_3... =(\mathrm{n}-1) \text { term ofthe sequence }\end{align} $6,10,18, \ldots$ which is a G.P. Adding column wise, we get \begin{align} & a_n-a_1=2+6+18+\ldots+(n-1) \text { terms } \\... } \\ & \Rightarrow a_n=3^{n-1}-1+3=3^{n-1}+2 \end{align} Taking summation of the both sides \begin{align}