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- Question 3 Exercise 5.3
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... \\ & 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 sequ
- Question 4 Exercise 5.3
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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 }\
- Question 5 Exercise 5.3
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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 sequenc
- Question 1 Exercise 5.3
- Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ===... 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
- Question 2 & 3 Exercise 5.4
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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
- Question 2 & 3 Exercise 5.2
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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
- Question 2 Exercise 5.3
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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) \
- Question 2 & 3 Exercise 5.1
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Q2... 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: $\quad T_j=j
- Question 7 Review Exercise
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 1^2, 2^2, 3^2, \ldots, n^2$. 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
- Question 6 Exercise 5.3
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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
- Question 2 & 3 Review Exercise
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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:so
- Question 5 & 6 Review Exercise
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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:so
- Question 8 Review Exercise
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... $ 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:so
- Question 9 Review Exercise
- Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... \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:so
- Question 1 Exercise 5.1
- Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... 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