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- Question 2 & 3 Exercise 5.1
- he sum $1.2+2.3+3.4+\ldots+99.100$. Solution: The given series is the product of the corresponding terms ... hose $n^{\text {th }}$ terms are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ term of the given series is: $\quad T_j=j(j+1)=j^2+j$. Taking sum o... +5^2+\ldots+99^2$. Solution: The each term of the given series is the square of the term of the series $1
- Question 1 Exercise 5.1
- ms. ====Solution==== We see that each term of the given series is square of the terms of the series $1+3+... rm is $2 n-1$. Therefore $n^{t h}$-term of - the given series is: $$T_j=(2 j-1)^2$$. Taking summation... +\ldots$ up to $n$ terms. ====Solution==== In the given series, we see that $T_1=1^2, T_2=1^2+2^2$, $T_3=
- Question 7 Review Exercise
- +5.4^2+\ldots$ to $n$ terms. ====Solution==== The given series if the product of corresponding terms of t... 3^2+\ldots$ to $n$ terms. ====Solution==== In the given series each term is the product of the correspond... 2$. Therefore, the $n^{\text {th }}$ term of the given series is: $$a_n=n^2 \cdot(2 n+1)\quad\text{or}\q
- Question 6 Exercise 5.1
- ms. ====Solution==== We see that each term of the given series is the product of corresponding terms of t... respectively, therefore the $n^{t h}$ term of the given series is: \begin{align} & T_j=j(j+1)(j+2)-j(j^2+
- Question 2 & 3 Exercise 5.2
- 4}{16}+\dfrac{5}{32}+\ldots$ ====Solution==== The given series is the product of the corresponding terms ... })^{n-1}$$ Thus the $n^{\text {th }}$ term of the given arithmetic-geometric series is: \begin{align} & T
- Question 9 Review Exercise
- g method of differences to compute the sum of the given series. \begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_... g method of differences to compute the sum of the given series. \begin{align} & a_2-a_1=5-2=3 \\ & a_3-a_
- Question 4 & 5 Exercise 5.1
- o $n$ terms. ====Solution==== First we reform the given series as: $$(1+1^2)+(1+2^2)+(1+3^2)+(1+4^2)+\ldo
- Question 9 Exercise 5.1
- ===Solution==== The $n$-term of the the series is given as: \begin{align} & T_n=n^2(2 n+3)=2 n^3+3 n^2 \\
- Question 4 & 5 Exercise 5.2
- \infty}&=3+\dfrac{2 r}{1-r}\end{align} But we are given $S_{\infty}=\dfrac{44}{9}$ \begin{align} & \there
- Question 2 & 3 Review Exercise
- .7+\ldots$ to $n$ terms. ====Solution==== In the given series each term is the product of corresponding
- Question 10 Review Exercise
- +\ldots$ ====Solution==== The general term of the given series is: \begin{align} a_n&=1+\dfrac{1}{2}+\dfr