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- Question 1 Exercise 5.2
- 2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \beg... +n \cdot 2^{n+1}-2^{n+1} \\ S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ te... x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align} Subtracting the (ii) from (i), we get \beg... (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 $n$ t
- Question 1 Exercise 5.3
- $$ Putting $A=1$,then \begin{align}1+B&=0\\ B&=-1\end{align} hence \begin{align} & \dfrac{1}{n(n+1)}=\d... \\ & \Rightarrow T_n=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking summation of the both sides of the ... +1}) \\ & =1-\dfrac{1}{n+1} \\ & =\dfrac{n}{n+1} \end{align} Hence the sum of the series is: $$S_n=\dfr... mathrm{I}=A(2 n+1)+B(2 n-1) \\ & =(2 A+2 B) n+A-B\end{align} Comparing coefficients of $n$ and constant
- Question 1 Exercise 5.1
- n[\dfrac{8 n^2-2}{6}] \\ & =\dfrac{n(4 n^2-1)}{3}\end{align} =====Question 1(ii)===== Sum the series $... 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 abo... 2}[n(n+1)+2(n+1)] \\ & =\dfrac{n(n+1)^2(n+2)}{12}\end{align} =====Question 1(iii)===== Sum the series ... (n+1)(2 n+1)}{6} \\ & =\dfrac{2 n(n+1)(2 n+1)}{3}\end{align} =====Question 1(iv)===== Sum the series $
- Question 8 Review Exercise
- \\ & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align} Thus the sum of $n$ terms is: $$S_n=\dfrac... dfrac{2(2n+1)+9}{6}\\ & =\dfrac{n(n+1)(4n+11)}{6}\end{align} Thus sum to $n$ terms is: $$S_n=\dfrac{n(n... +4) \\ & a_n==n(n^2+5 n+4) \\ & a_n=n^3+5 n^2+4 n\end{align} Taking summation of the both sides \begin{... 0+24}{6}] \\ & =\dfrac{n(n+1)(3 n^2+23 n+34)}{12}\end{align} Thus the sum to $n$ terms is: $$S_n=\dfrac
- Question 9 Review Exercise
- \text { term of the series } \\ & 4,6,8, \ldots \end{align} Adding column wise, we get \begin{align} a... +n-2+3 \because a_1=3 \\ \Rightarrow a_n&=n^2+n+1\end{align} Taking summation of the both sides \begin{... 3 n+3+6}{6}] \\ & =n \cdot \dfrac{2 n^2+6 n+9}{6}\end{align} Thus the sum to $n$ terms is: $$S_n=\dfrac... 1=5-2=3 \\ & a_3-a_2=14-5=9 \\ & a_4-a_3=41-14=27\end{align} $a_n-a_{n-1}=(n-1)$ term of the series $3,
- Question 4 & 5 Exercise 5.2
- {3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align} Subtracting the (ii) from (i), we get \beg... s) \\ & =5+2 \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}\end{align} $\because$ the series is geometric $r=\dfr... {2}=\dfrac{18}{2}=9 \\ & \Rightarrow S_{\infty}=9\end{align} =====Question 5===== If the sum to infini... -r} \\ \Rightarrow S_{\infty}&=3+\dfrac{2 r}{1-r}\end{align} But we are given $S_{\infty}=\dfrac{44}{9}
- Question 2 & 3 Exercise 5.1
- =\frac{99(99+1)(2(99)+1)}{6}+\frac{99(99+1)}{2} \end{aligned} $$ $$ \begin{aligned} & =\frac{99(100)(... tarrow 1.2+2.3+3.4+\ldots+99.100 \\ & =3368050 . \end{aligned} $$ Q3 Find the sum $1^2+3^2+5^2+\ldots+... w 2 j-1=99 \\ & \Rightarrow j=\frac{100}{2}=50 . \end{aligned} $$ The sum of the 50 terms of the series... ightarrow 1^2+3^2+5^2+ \\ & \ldots+99^2=156850 . \end{aligned} $$ ====Go To==== <text align="left"><bt
- Question 7 & 8 Exercise 5.1
- in{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align} Taking sum of the both sides of the above ... [n(n+9)+8(n+9)] \\ & =\dfrac{n(n+1)(n+8)(n+9)}{4}\end{align} =====Question 8===== Find the sum of the ... n[4 \dfrac{8 n^2+6 n+1}{6}+\dfrac{5(2 n+1)}{2}+1]\end{align} \begin{align} & =2 n[\dfrac{32 n^2+24 n+4+... 30 n+15+6}{6}] \\ & =\dfrac{n}{3}(32 n^2+54 n+25)\end{align} ====Go To==== <text align="left"><btn
- Question 2 & 3 Exercise 5.2
- S_{\infty}=x+3^2 x^2+5^2 x^3+7^2 x^4+\ldots..(2)\end{align} Subtracting the (2) from (2), we get \begi... x(1-x) S_{\infty}=x+8 x^2+16 x^3+24 x^4+\ldots(4)\end{align} Again subtracting (4) from (3) \begin{alig... ty}=\dfrac{1+7 x}{(x-1)^2}-\dfrac{8x^2}{(x-1)^3} \end{align} =====Question 3===== Find the $n^{\text {... \times b_n \\ & T_n=(n)\cdot(\dfrac{1}{2})^{n-1}\end{align} ====Go To==== <text align="left"><btn ty
- Question 2 & 3 Exercise 5.4
- } \\ S_n&=\sum_{k=1}^n \dfrac{1}{(3 k-1)(3 k-2)} \end{align} The $n$ term of the above series is: $$u_n... k+2)+B(3 k-1) \\ & \Rightarrow(3 A+3 B) k+2 A-B=1\end{align} Comparing the cocfficients of $k$ and cons... }{2}-\dfrac{1}{3 n+2}]\\ S_n&=\dfrac{n}{2(3 n+2)}\end{align} =====Question 3===== Find the sum of the ... dfrac{1}{3})+\ldots+(\dfrac{1}{n-1}-\dfrac{1}{n})\end{align} Hence the sum is: $$S_n=1-\dfrac{1}{n}=\df
- Question 4 Review Exercise
- dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align} Multiplying both sides by $(3 n-2)(3 n+1)(... A+9 B+9 C] n^2+[15 A+6 B-3 C] n+[4 A+8 B-2 C]&=1\end{align} Comparing the coefficients of $n \cdot n$ ... frac{1}{3 n-2}-\dfrac{2}{3 n+1}+\dfrac{1}{3 n+4}]\end{align} Taking summation of the both sides \begin{... 1-\dfrac{1}{4}-\dfrac{1}{3 n+1}+\dfrac{1}{3 n+4}]\end{align} Thus the sum of $n$ term is: $$S_n=\dfrac{
- Question 5 & 6 Review Exercise
- 19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align} Subtracting the (ii) from (i) we get \begi... ac{7(x-x^n)}{(1-x)^2}-\dfrac{(7 n-1) x^n}{1-x}\\ \end{align} =====Question 6===== Sum the series: $\df... \dfrac{1}{n+1}\\ T_n&=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking summation of the both sides of the ... c{1}{n+1}) \\ & =1-\dfrac{1}{n+1}=\dfrac{n}{n+1} \end{align} Hence the sum of the series is: $$S_n=\dfr
- Question 4 & 5 Exercise 5.1
- &=\dfrac{j(3 j+1)}{2} \\ & =\dfrac{1}{2}(3 j^2+j)\end{align} Taking sum of the both sides of the above ... dfrac{n(n+1)}{4}[2 n+2] \\ & =\dfrac{n^2}{2}(n+1)\end{align} =====Question 5===== Sum: $2+5+10+17+\ldo... +2 n^2+3 n+1}{6}] \\ & =\dfrac{n}{6}(2 n^2+3 n+7)\end{align} ====Go To==== <text align="left"><btn
- Question 9 Exercise 5.1
- n+3)=2 n^3+3 n^2 \\ & \Rightarrow T_j=2 j^3+3 j^2\end{align} Taking sum of the both sides of the above ... }[n^2+n+2 n+1] \\ & =\dfrac{n(n+1)}{2}(n^2+3 n+1)\end{align} =====Question 9(ii)===== Find the sum of ... +n(n+1)(-n^2+ n+1)\\ &=4^{n+1}-4-n(n+1)(n^2- n-1)\end{align} ====Go To==== <text align="left"><btn t
- Question 1 Exercise 5.3
- )th \quad\text{term of sequence}\quad 9,15,21,...\end{align} Which is in A.P. column wise, we get \begi... -3+4 \because a_1=4 \\ \Rightarrow a_n&=3 n^2+1 \end{align} Taking summation of the both sides \begin{... t{Hence} \quad 3 n^2+1&;\dfrac{n}{2}(2 n^2+3 n+3)\end{align} ====Go To==== <text align="right"><btn