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Question 1 Exercise 5.3
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$n$ and constants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,th... } Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfr... $n$ and constants on the both sides of the above equation, we get $$3 A+3 B=0 \quad \text{and} \quad 2 A-B... $n$ and constants on the both sides of the above equation, we get $$9 A+9 B=0\quad \text{ and} \quad 4 A-5
Question 1 Exercise 5.1
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Taking summation of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1... {align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=64 \sum_
Question 4 & 5 Exercise 5.1
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{align} Taking sum of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_i=\dfrac{1}... 1+j^2$ Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^{j=n} T_j=\sum
Question 9 Exercise 5.1
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{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=2 \sum_{... 4 n^3$$ Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=3 \sum_{
Question 2 & 3 Exercise 5.4
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$k$ and constants on the both sides of the above equation, we get $$3 A+3 B=0\quad \text{and}\quad 2 A-B=1... }=\dfrac{A}{n}+\dfrac{B}{n-1}$$ Solving the above equation for $A$ and $B$ we get $A=1$ and $B=-1$. So, \be
Question 5 & 6 Review Exercise
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$n$ and constants on the both sides of the above equation, we get $$A+B=0\quad \text{and}\quad A=1$$ Puttin... } Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfr
Question 6 Exercise 5.1
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{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=
Question 7 & 8 Exercise 5.1
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{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=
Question 4 Exercise 5.4
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\dfrac{A}{n+3}+\dfrac{B}{n+4}$$ Solving the above equation for $A$ and $B$, we get $A=1$ and $B=-1$, so $$u