Search
You can find the results of your search below.
Fulltext results:
- Question 1 Exercise 5.3
- $$u_n=\dfrac{1}{(3 n-1)(3 n+2)}$$ Resolving into partial fractions: $$\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{A}{3 n-1}-\dfrac{B}{3 n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Rightarrow(3 A+3 B) n+2 A-B&=1\end{align} Comparing the coefficients of $n$ and consta
- Question 4 Review Exercise
- neral term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partial fractions \begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align} Multiplying both side
- Question 2 & 3 Exercise 5.4
- ====== Question 2 & 3 Exercise 5.4 ====== Solutions of Question 2 & 3 of Exercise 5.4 of Unit 05: Miscullaneous Series... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2===== Find sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let
- Question 5 & 6 Review Exercise
- kistan. =====Question 5===== Sum the series: $5+12 x+19 x^2+26 x^3+\ldots$ to $n$ terms. ====Solution==== Let \begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}...(i)\\ x
- Question 4 Exercise 5.4
- the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$ ====Solution==== Let \begin{align}S_n &=\sum_{k=1}^n \dfrac{1}{k^2+7 k+12} \\ & =\sum_{k=1}^n \dfrac{1}{(k+3)(k+4)}\end{align} Consider the $n^{\text {th }}$ term of t