Search
You can find the results of your search below.
Fulltext results:
- Question 1 Exercise 5.2
- +2.2^2+3.2^3+4.2^4+\ldots$. ====Solution==== Let \begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4+\l... end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(... 1+4 x+7 x^2+10 x^3+\ldots.$ ====Solution==== Let \begin{align} & S_n=1+4 x+7 x^2+10 x^3+\ldots +(3 n-2) x... end{align} Subtracting the (ii) from (i), we get \begin{align} (1-x) S_n&=1+(4-1) x+(7-4) x^2+ (10-7) x^3
- Question 1 Exercise 5.3
- get $$A+B=0 \text{and} A=1$$ 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} \\ &... ummation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfrac{1}{k... tiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm{I}=A(2 n+1)+B(2 n-1) \\ & =(2 A+
- Question 1 Exercise 5.1
- of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2 \\... T_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... ng of sum of the both sides of the above, we get \begin{align}\Rightarrow \sum_{j=1}^n T_j&=\dfrac{1}{6} ... )^2=4 j^2$ Taking sum of the both sides, we get \begin{align}& \sum_{j=1}^n T_j=4 \sum_{j=1}^{j=n} j^2 \
- Question 8 Review Exercise
- a_n=n^3+3^n$$ Taking summation of the both sides \begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_{r=... n=2 n^2+3 n$$ Taking summation on the both sides \begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^2+3 \su... ====Solution==== The $n^{\text {th }}$ term is: \begin{align} & a_n=n(n+1)(n+4) \\ & a_n==n(n^2+5 n+4) \... n\end{align} Taking summation of the both sides \begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+5 \sum_
- Question 9 Review Exercise
- ferences 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... 8, \ldots \end{align} Adding column wise, we get \begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & =\dfrac{n-1... +1\end{align} Taking summation of the both sides \begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r... ferences to compute the sum of the given series. \begin{align} & a_2-a_1=5-2=3 \\ & a_3-a_2=14-5=9 \\ & a
- Question 4 & 5 Exercise 5.2
- ^2}+\dfrac{11}{3^3}+\ldots$ ====Solution==== Let \begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}... end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2}{3}+\... ause$ the series is geometric $r=\dfrac{1}{3}<1$ \begin{align}& =5+2 \cdot \dfrac{1}{3} \cdot \dfrac{3}{2... ..(ii)$$ Subtracsing the (ii) from (i), we get \begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2 r^3+\ldots
- Question 2 & 3 Exercise 5.1
- f the both sides from $j=1$ to $j=99$, we get $$ \begin{aligned} & \sum_{j=1}^{99} \tau_j=\sum_{j=1}^{99}... )+1)}{6}+\frac{99(99+1)}{2} \end{aligned} $$ $$ \begin{aligned} & =\frac{99(100)(199)}{6}+\frac{99(100)}... total number of terms in the given series as: $$ \begin{aligned} & a_j=99=1+2(j-1) \\ & \Rightarrow 2 j-1... ned} $$ The sum of the 50 terms of the series $$ \begin{aligned} & \sum_{j=1}^{j=50} T_j=\sum_{j=1}^{j=50
- Question 7 & 8 Exercise 5.1
- general term of the series is: $T_j=j(j+4)(j+8)$ \begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\en... of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \su... king summation of the general term of the series \begin{align} & \sum_{j=1}^{2 n} T_j=4 \sum_{j=1}^{2 n} ... n^2+6 n+1}{6}+\dfrac{5(2 n+1)}{2}+1]\end{align} \begin{align} & =2 n[\dfrac{32 n^2+24 n+4+30 n+15+6}{6}]
- Question 2 & 3 Exercise 5.2
- 2 x^2+7^2 x^3+\ldots, x<1$. ====Solution==== Let \begin{align} & S_{\infty}=1+3^2 x+5^2 x^2+7^2 x^3+\ldot... \end{align} Subtracting the (2) from (2), we get \begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2) x+(5^2-3^2... ots(4)\end{align} Again subtracting (4) from (3) \begin{align} & [(1-x)-x(1-x)] S_{\infty}=1+(8-1) x+(16-... erm of the given arithmetic-geometric series is: \begin{align} & T_n=a_n \times b_n \\ & T_n=(n)\cdot(\df
- Question 2 & 3 Exercise 5.4
- {k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9... -1}+\dfrac{B}{3 k+2}$$ Multiplying both sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & 1=A... {1}{3 k+2}]$$ Taking summation of the both sides \begin{align} \sum_{r=1}^n u_n&=\sum_{r=1}^n[\dfrac{1}{3... on for $A$ and $B$ we get $A=1$ and $B=-1$. So, \begin{align} u_n&=\dfrac{1}{n}-\dfrac{1}{n-1} \\ S_n&=\
- Question 4 Review Exercise
- n+1)(3 n+4)}$$ Resolving into partial fractions \begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A... ng 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)... dfrac{1}{9}$ and $C=\dfrac{1}{18}$. Thus we have \begin{align} & a_n=\dfrac{1}{18(3 n-2)}-\dfrac{1}{9(3 n... }]\end{align} Taking summation of the both sides \begin{align}& \sum_{r=1}^n a_r \\ & =\dfrac{1}{18} \sum
- Question 5 & 6 Review Exercise
- 6 x^3+\ldots$ to $n$ terms. ====Solution==== Let \begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^... \end{align} Subtracting the (ii) from (i) we get \begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\ ... $$ Putting $A=1$ in the $1+B=0$, we get $$B=-1$$ \begin{align}\dfrac{1}{n(n+1)}&=\dfrac{1}{n}-\dfrac{1}{n... ummation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfrac{1}{k
- Question 4 & 5 Exercise 5.1
- olution==== The general term of the sequence is: \begin{align}& T_j=\dfrac{j}{2}[2(2)+3(j-1)]\\ &=\dfrac{... 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... of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^{j=n} T_j=\sum_{j=1}^{j=n} 1+
- Question 9 Exercise 5.1
- ==== The $n$-term of the the series is given as: \begin{align} & T_n=n^2(2 n+3)=2 n^3+3 n^2 \\ & \Rightar... of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=2 \sum_{j=1}^n j^3+3 \s... of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=3 \sum_{j=1}^n 4^j+6 \s
- Question 1 Exercise 5.3
- e series $4+13+28+49+76+\ldots$ ====Solution==== \begin{align} & a_2-a_1=13-4=9 \\ & a_3-a_2=28-13=15 \\ ... \end{align} Which is in A.P. column wise, we get \begin{align} a_n-a_{n-1}&=9+15+21+\ldots+(n-1)\\ & =\df... 1 \end{align} Taking summation of the both sides \begin{align} & \sum_{r=1}^n a_r=3 \sum_{r=1}^n r^2+\sum