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- Question 1 Exercise 5.1
- stion 1 of Exercise 5.1 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Gr... hawar, Pakistan. =====Question 1(i)===== Sum the series $1^2+3^2+5^2+7^2+\ldots$ up to $n$ terms. ====Solution==== We see that each term of the given series is square of the terms of the series $1+3+5+\ldots$ whose $n^{\text {th }}$ term is $2 n-1$. Therefore
- Question 1 Exercise 5.3
- stion 1 of Exercise 5.4 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Gr... istan. =====Question 1(i)==== Find the sum of the series $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ld... $ terms. ====Solution==== The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ int... =\dfrac{n}{n+1} \end{align} Hence the sum of the series is: $$S_n=\dfrac{n}{n+1}$$ =====Question 1(ii)==
- Question 7 Review Exercise
- on 7 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... tan. =====Question 7(i)===== Find the sum of the series: $1.2^2+3.3^2+5.4^2+\ldots$ to $n$ terms. ====Solution==== The given series if the product of corresponding terms of the two series $1,3,5, \ldots,(2 n-1)$ and $2^2, 3^2, 4^2, \ldot
- Question 2 & 3 Exercise 5.1
- n 2 & 3 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... $1.2+2.3+3.4+\ldots+99.100$. Solution: The given series is the product of the corresponding terms of the series $1+2+3+\ldots+99$ and $2+3+4+\ldots+100$, whose $... n^{\text {th }}$ terms are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ t
- Question 9 Review Exercise
- on 9 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... )===== Find the sum of the first $n$ terms of the series $3+7+13+21+31+\ldots$ ====Solution==== Using meth... od of differences to compute the sum of the given series. \begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=... \ldots \\ & a_n-a_{n-1}=(n-1) \text { term of the series } \\ & 4,6,8, \ldots \end{align} Adding column wi
- Question 2 & 3 Exercise 5.2
- n 2 & 3 of Exercise 5.2 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... th }}$ term of the following arithmetic-geometric series: $\dfrac{0}{1}+\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3... +\dfrac{5}{32}+\ldots$ ====Solution==== The given series is the product of the corresponding terms of the series: $0,1.2,3, \ldots$ and $1, \dfrac{1}{2}, \dfrac{1
- Question 1 Exercise 5.2
- stion 1 of Exercise 5.2 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Gr... =====Question 1(i)===== Sum up to $n$ terms the series $1.2+2.2^2+3.2^3+4.2^4+\ldots$. ====Solution==== ... =====Question 1(ii)===== Sum up to $n$ terms the series $1+4 x+7 x^2+10 x^3+\ldots.$ ====Solution==== Let... =====Question 1(iii)===== Sum up to $n$ terms the series $1+2 x+3 x^2+4 x^3+\ldots$. ====Solution==== Let
- Question 1 Review Exercise 5
- n+11$</collapse> ii. The sum to infinity of the series: $1+\dfrac{2}{3}+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\... llapsed="true">(c): $3$ </collapse> iii. Sum the series:$1+2.2+3.2^2+\cdots+100.2^{\prime \prime}$ * ... 100}+1$</collapse> iv. The $n^{t h}$ term of the series: $1.2+2.3+3.4+\ldots$ * (a) $n^2-n$ * %%... : $n^2+n$ </collapse> v. Sum of $n$ terms of the series whose $n^{t h}$ term is $1+2^n$ * (a) $n \
- Question 2 & 3 Review Exercise
- & 3 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... Peshawar, Pakistan. =====Question 2===== Sum the series to $n$ terms $1.2+2.3+3.4+\ldots$ ====Solution===... n+2)}{3}\end{align} =====Question 3===== Sum the series: $1.3 .5+2.4 .6+3.5 .7+\ldots$ to $n$ terms. ====Solution==== In the given series each term is the product of corresponding terms o
- Question 9 Exercise 5.1
- stion 9 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... on 9(i)===== Find the sum of the $n$ terms of the series whose $n$-term is (32 n^2+54 n+25). ====Solution==== The $n$-term of the the series is given as: \begin{align} & T_n=n^2(2 n+3)=2 n^3... n 9(ii)===== Find the sum of the $n$ terms of the series whose $n$-term is $3\left(4^n+2 n\right)-4 n^3$.
- Question 2 & 3 Exercise 5.4
- n 2 & 3 of Exercise 5.4 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... , Pakistan. =====Question 2===== Find sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Soluti... -1)(3 k-2)} \end{align} The $n$ term of the above series is: $$u_n=\dfrac{1}{(3 k-1)(3 k+2)}$$ Resolving i... {align} =====Question 3===== Find the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2-k}$ ====Solution====
- Question 5 & 6 Review Exercise
- & 6 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... Peshawar, Pakistan. =====Question 5===== Sum the series: $5+12 x+19 x^2+26 x^3+\ldots$ to $n$ terms. ====... {1-x}\\ \end{align} =====Question 6===== Sum the series: $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\l... ===Solution==== Solution: The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ int
- Question 8 Review Exercise
- on 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... estion 8(i)===== Find the sum of $n$ terms of the series whose $n^{t h}$ term is $n^3+3^n.$ ====Solution==... stion 8(ii)===== Find the sum of $n$ terms of the series whose $n^{t h}$ term is $2 n^2+3 n$ ====Solution=... tion 8(iii)===== Find the sum of $n$ terms of the series whose $n^{t h}$ term is $n(n+1)(n+4)$ ====Solutio
- Question 6 Exercise 5.1
- stion 6 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... ==Solution==== We see that each term of the given series is the product of corresponding terms of the three series $1+2+3+\ldots, \quad 2+3+4+5+\ldots$ and $3+4+5+6... tively, therefore the $n^{t h}$ term of the given series is: \begin{align} & T_j=j(j+1)(j+2)-j(j^2+3 j+2)
- Question 7 & 8 Exercise 5.1
- n 7 & 8 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Gr... +\ldots$ ====Solution==== The general term of the series is: $T_j=j(j+4)(j+8)$ \begin{align} & =j(j^2+12 j... {align} =====Question 8===== Find the sum of the series upto $2 n$ terms, whose $n^{t h}$-term is $4 n^2+... n==== Taking summation of the general term of the series \begin{align} & \sum_{j=1}^{2 n} T_j=4 \sum_{j=1}