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- Question 3 Exercise 5.3
- \\ & a_3-a_2=18-10=8 \\ & a_4-a_3=28-18=10 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n \quad 1}=(\mathrm{n}-1) \text { term of the sequence } \end{align} $6,10,8, \ld... get \begin{align}& a_n-a_1=6+10+8-\ldots +(n-1) \text { terms } \\ & =\dfrac{n-1}{2}[2 \cdot 6+(n-2) \c
- Question 4 Exercise 5.3
- 2 \\ & a_3-a_2=11-5=6 \\ & a_4-a_3=29-11=18 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence }\end{align} $6,10,18, \ldot... get \begin{align} & a_n-a_1=2+6+18+\ldots+(n-1) \text { terms } \\ & =\dfrac{2 \cdot[3^{n- 1}-1]}{3-1}
- Question 5 Exercise 5.3
- 6 \\ & a_3-a_2=21-9=12 \\ & a_4-a_3=45-21=24\\ & \text {... ... ... } \\ & \text {... ... ... } \\ &a_n-a_{n-1}=(\mathrm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{a... get \begin{align} a_n-a_1& =6+12+24+\ldots+(n-1) \text {terms } \\ & =\dfrac{6[2^{n-1}-1]}{2-1} \\ \Righ
- Question 1 Exercise 5.3
- both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \begin{align}1+B&=0... oefficients of $n$ and constant term $$2 A+2 B=0 \text { and } A-B=1 $$ Solving the above two equations ... $ and $B$, we get \begin{align}A&=\dfrac{1}{2}\\ \text{and} B&=-\dfrac{1}{2}\end{align} Hence $$u_n=\dfr... of the above equation, we get $$3 A+3 B=0 \quad \text{and} \quad 2 A-B=1$$ Solving the above two equati
- Question 2 & 3 Exercise 5.4
- c{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2} ... ing both sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & 1=A(3 k+2)+B(3 k-1) \\ & \Rightar... of the above equation, we get $$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=-\dfrac{1}{3}$$ Thus $$u_n=\dfrac{1}{
- Question 2 & 3 Exercise 5.2
- 1-x) S_{\infty}=1+8 x+16 x^2+24 x^3+\ldots \quad \text { (3) } \\ & \Rightarrow x(1-x) S_{\infty}=x+8 x^... } \end{align} =====Question 3===== Find the $n^{\text {th }}$ term of the following arithmetic-geometri... ies is $$b_n=(\dfrac{1}{2})^{n-1}$$ Thus the $n^{\text {th }}$ term of the given arithmetic-geometric se... t(\dfrac{1}{2})^{n-1}\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 2 Exercise 5.3
- dots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,16,22, \ldots\end{align... get \begin{align} a_n-a_1&=10+16+22+\ldots+(n-1) \text { terms } \\ & =\dfrac{n-1}{2}[2 \cdot 10+(n-2) \... {2} \\ & \Rightarrow \sum_{r=1}^n a_r=n(n+1)^2\\ \text{Hence}\quad 3 n^2+n&;n(n+1)^2\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 2 & 3 Exercise 5.1
- 2+3+\ldots+99$ and $2+3+4+\ldots+100$, whose $n^{\text {th }}$ terms are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ term of the given series is: $\quad T_j=j... ts+99^2=156850 . \end{aligned} $$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:ex5-1-p1 |< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 7 Review Exercise
- 1^2, 2^2, 3^2, \ldots, n^2$. Therefore, the $n^{\text {th }}$ term of the given series is: $$a_n=n^2 \cdot(2 n+1)\quad\text{or}\quad a_n=2 n^3+n^2$$ Taking summation of the ... frac{n(n+1)(3 n^2+5 n+1)}{6}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:Re-ex5-p4 |< Question 5 & 6 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 6 Exercise 5.3
- ots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence } 4,20,100, \ldots \end{ali... get \begin{align} a_n-a_1&=4+20+100+\ldots+(n-1) \text { terms } \\ & =\dfrac{4[5^{n -1}-1]}{5-1} \\ \Ri... rrow \sum_{r=1}^n a_r&=\dfrac{(5^n-1)}{4}+27 n\\ \text{Hence}\quad5^{n-1}+27; \dfrac{(5^n-1)}{4}+27 n\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 2 & 3 Review Exercise
- ms $1.2+2.3+3.4+\ldots$ ====Solution==== The $n^{\text {th }}$ term is: $$a_n=n(n+1)=n^2+n$$ Taking summ... \dfrac{n(n+1)(n+4)(n+5)}{4}$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:Re-ex5-p1 |< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit05:Re-ex5-p3|Question 4 >]]</btn></text
- Question 5 & 6 Review Exercise
- 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 ... ries is: $$S_n=\dfrac{n}{n+1} $$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:Re-ex5-p3 |< Question 4 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit05:Re-ex5-p5|Question 7 >]]</btn></text
- Question 8 Review Exercise
- $ term is $n(n+1)(n+4)$ ====Solution==== The $n^{\text {th }}$ term is: \begin{align} & a_n=n(n+1)(n+4) ... s: $$S_n=\dfrac{n(2n^2+1)}{3}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:Re-ex5-p5 |< Question 7 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit05:Re-ex5-p7|Question 9 >]]</btn></text
- Question 9 Review Exercise
- \quad \cdots \quad \ldots \\ & a_n-a_{n-1}=(n-1) \text { term of the series } \\ & 4,6,8, \ldots \end{al... c{3}{4}(3^n-1)+\dfrac{3 n}{2}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:Re-ex5-p6 |< Question 8 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit05:Re-ex5-p8|Question 10 >]]</btn></tex
- Question 1 Exercise 5.1
- he terms of the series $1+3+5+\ldots$ whose $n^{\text {th }}$ term is $2 n-1$. Therefore $n^{t h}$-ter... ch term of the series $1+5+9+\ldots$, whose $n^{\text {th }}$ term is: $a_j=1+4(j-1)=4 j-3$. \begin{ali... 6 n^3-16 n^2-2 n+3]\end{align} ====Go To==== <text align="right"><btn type="success">[[math-11-kpk:sol:unit05:ex5-1-p3|Question 2 & 3 >]]</btn></text>