Search
You can find the results of your search below.
Fulltext results:
- Question 1 Exercise 5.3
- onstants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \... 1)}$$ Multiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm{I}=A(2 n+1)+B(2 n-1)... Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1}{2}\\ \text{and} B&=... n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Right
- Question 1 Exercise 5.1
- 5^2+7^2+\ldots$ up to $n$ terms. ====Solution==== We see that each term of the given series is square ... ummation of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(... $n$ terms. ====Solution==== In the given series, we see that $T_1=1^2, T_2=1^2+2^2$, $T_3=1^2+2^2+3^2$ and so on we get \begin{align}& T_j=1^2+2^2+3^2+\ldots+j^2 \\
- Question 1 Exercise 5.2
- ...(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+... ...(ii)\end{align} Subtracting the (ii) from (i), we get \begin{align} (1-x) S_n&=1+(4-1) x+(7-4) x^2+... ..(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align}& (1-x) S_n=1+(2-1) x+(3-2) x^2 ... align} Multiply (i) both sides by $\dfrac{1}{2}$, we get $$\dfrac{1}{2} S_n=\dfrac{1}{2}+\dfrac{3}{2^2
- Question 4 Review Exercise
- ultiplying both sides by $(3 n-2)(3 n+1)(3 n+4)$, we get \begin{align} 1=A(3 n+1)(3 n+4)+B(3 n-2)(3 n+... nd constant terms on the both sides of equations, we get $$A+B+C=0 \quad 15 A+6 B-3 C=0$$ $$4 A+8 B-2 ... three equations for the constants $A, B$ and $C$ we get $A=\dfrac{1}{18}, B=-\dfrac{1}{9}$ and $C=\dfrac{1}{18}$. Thus we have \begin{align} & a_n=\dfrac{1}{18(3 n-2)}-\df
- Question 2 & 3 Exercise 5.4
- h sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & 1=A(3 k+2)+B(3 k-1) \\ & \Rightarrow(3... onstants on the both sides of the above equation, we get $$3 A+3 B=0\quad \text{and}\quad 2 A-B=1$$ Solving the above two equations for $A$ and $B$ we get $$A=\dfrac{1}{3}\quad\text{and}\quad B=-\dfra... -1}$$ Solving the above equation for $A$ and $B$ we get $A=1$ and $B=-1$. So, \begin{align} u_n&=\dfr
- Question 5 & 6 Review Exercise
- ....(ii)\end{align} Subtracting the (ii) from (i) we get \begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^... c{B}{(n+1)}$$ Multiplying both sides by $n(n+1)$, we get $$1=A(n+1)+B n=(A+B) n+A$$ Comparing the coef... onstants on the both sides of the above equation, we get $$A+B=0\quad \text{and}\quad A=1$$ Putting $A=1$ in the $1+B=0$, we get $$B=-1$$ \begin{align}\dfrac{1}{n(n+1)}&=\dfr
- Question 4 & 5 Exercise 5.1
- king sum of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_i=\dfrac{1}{2}[... 0+17+\ldots$ to $n$ terms. ====Solution==== First we reform the given series as: $$(1+1^2)+(1+2^2)+(1+... king sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^{j=n} T_j=\sum_{j=
- Question 4 & 5 Exercise 5.2
- s.(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\df... \infty....(ii)$$ Subtracsing the (ii) from (i), we get \begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2... row S_{\infty}&=3+\dfrac{2 r}{1-r}\end{align} But we are given $S_{\infty}=\dfrac{44}{9}$ \begin{align
- Question 2 & 3 Exercise 5.1
- aking sum of the both sides from $j=1$ to $j=99$, we get $$ \begin{aligned} & \sum_{j=1}^{99} \tau_j=\... e term of the series $1+3+5+\ldots+99$. So, first we find the total number of terms in the given serie
- Question 6 Exercise 5.1
- 3.4+3.4 .5+\ldots$ to $n$-terms. ====Solution==== We see that each term of the given series is the pro... king sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n
- Question 9 Exercise 5.1
- king sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=2 \sum_{j=1}... king sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=3 \sum_{j=1}
- Question 3 Exercise 5.3
- he series $4+10+18+28+40+\ldots$ ====Solution==== We use the method of difference as: \begin{align} & ... 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) \
- Question 9 Review Exercise
- \ & 4,6,8, \ldots \end{align} Adding column wise, we get \begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & ... the series $3,9,27, \ldots$ Adding column wise, we get $$a_n-a_1=3+9+27+\cdots+3^n$$ Which is a geom
- Question 7 & 8 Exercise 5.1
- king sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n
- Question 2 & 3 Exercise 5.2
- ots..(2)\end{align} Subtracting the (2) from (2), we get \begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2)