Search
You can find the results of your search below.
Fulltext results:
- Question 1 Exercise 5.1
- \dfrac{n(n+1)}{2}+n \\ & =n[\dfrac{4(n+1)(2 n+1)-12(n+1)+6}{6}] \\ & =n[\dfrac{4(2 n^2+3 n+1)-12 n-12+6}{6}] \\ & =n[\dfrac{8 n^2+12n+4-12 n-6}{6}] \\ & =n[\dfrac{8 n^2-2}{6}] \\ & =\dfrac{n(4 n^2-1)}{3}
- Question 8 Review Exercise
- n+10+24}{6}] \\ & =\dfrac{n(n+1)(3 n^2+23 n+34)}{12}\end{align} Thus the sum to $n$ terms is: $$S_n=\dfrac{n(n+1)(3 n^2+23 n+34)}{12}$$ =====Question 8(iv)===== Find the sum of $n$ ... }-2]+n \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{8 n+4-12}{6}]+n \\ & =\dfrac{n(n+1)(8 n-8)}{12}+n \\ & =\dfrac{4n(n+1)(2 n-2)}{12}+n \\ & =\dfrac{4(n(2n^2-2n+
- Question 5 & 6 Review Exercise
- akistan. =====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 S_n&=5 x+12 x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....... e (ii) from (i) we get \begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\ &+[7 n-2-(7 n-9)] x^{n-
- Question 6 Exercise 5.1
- 3}{3}+2] \\ & =\dfrac{n(n+1)}{2}[\dfrac{3 n^2+3 n+12 n+6+12}{6}] \\ & =\dfrac{n(n+1)}{12}[3 n^2+15 n+18] \\ & =\dfrac{3 n(n+1)}{12}[n^2+5 n+6] \\ & =\dfrac{n(n+1)(n^2+5 n+6)}{4} \\ &=\dfrac{
- Question 7 & 8 Exercise 5.1
- ries is: $T_j=j(j+4)(j+8)$ \begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \sum_{j=1}^n j^2+32 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+12 \dfrac{n(n+1)(2 n+1)}{6}+32 \dfrac{n(n+1)}{2} \\
- Question 5 Exercise 5.3
- = \begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21-9=12 \\ & a_4-a_3=45-21=24\\ & \text {... ... ... } \\... rm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{align} which is a G.P. Adding column wise, we get \begin{align} a_n-a_1& =6+12+24+\ldots+(n-1) \text {terms } \\ & =\dfrac{6[2^{
- Question 4 Exercise 5.4
- um 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{ali
- Question 1 Exercise 5.3
- }[2.9+(n-2) \cdot 6] \\ & =\dfrac{n-1}{2}[18+6 n-12] \\ & =\dfrac{n-1}{2}[6+6 n] \\ \Rightarrow a_n&
- Question 2 Exercise 5.3
- dot 10+(n-2) \cdot 6] \\ & =\dfrac{n-1}{2}[20+6 n-12] \\ & =\dfrac{n-1}{2}[6 n+8] \\ & =2 \cdot \dfrac
- Question 3 Exercise 5.3
- {2}[2 \cdot 6+(n-2) \cdot 2] \\ & =\dfrac{n-1}{2}[12+2 n-4] \\ & =\dfrac{n-1}{2}[2 n+8]=2 \cdot \dfrac