Search
You can find the results of your search below.
Fulltext results:
- Question 1 Exercise 5.2
- n=1.2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) ... S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ terms the series $1+4 x+7 x^2... =x+4 x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align} Subtracting the (ii) from (i), we get
- Question 4 & 5 Exercise 5.2
- ac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2... $$r S_{\infty}=3 r+5 r^2+7 r^3+\ldots \infty....(ii)$$ Subtracsing the (ii) from (i), we get \begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2 r^3+\ldots \
- Question 1 Review Exercise 5
- d="a1" collapsed="true">(b): $6 n+11$</collapse> ii. The sum to infinity of the series: $1+\dfrac{2}{... e id="a2" collapsed="true">(c): $3$ </collapse> iii. Sum the series:$1+2.2+3.2^2+\cdots+100.2^{\prime... id="a6" collapsed="true">(c): $2076$</collapse> vii. What is the $n$ term of the series: $1+\dfrac{1+... lapsed="true">(a): $\dfrac{n+1}{2}$</collapse> viii. Sum of $n$ terms of the series $1^3+3^3+5^3+7^3+
- Question 1 Exercise 5.1
- \dfrac{n(4 n^2-1)}{3}\end{align} =====Question 1(ii)===== Sum the series $1^2+(1^2+2^2)+(1^2+2^2+3^2)... c{n(n+1)^2(n+2)}{12}\end{align} =====Question 1(iii)===== Sum the series $2^2+4^2+6^2+8^2+\ldots$ up
- Question 1 Exercise 5.3
- eries is: $$S_n=\dfrac{n}{n+1}$$ =====Question 1(ii)==== Find the sum of the series $\dfrac{1}{1.3}+\... _n&=\dfrac{n}{2 n+1}\end{align} =====Question 1(iii)==== Find the sum of the series $\dfrac{1}{2.5}+\
- Question 5 & 6 Review Exercise
- x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align} Subtracting the (ii) from (i) we get \begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdot
- Question 8 Review Exercise
- n+1)^2}{4}+\dfrac{3}{2}(3^n-1)$$ =====Question 8(ii)===== Find the sum of $n$ terms of the series who... $S_n=\dfrac{n(n+1)(4n+11)}{6}$$ =====Question 8(iii)===== Find the sum of $n$ terms of the series who
- Question 9 Exercise 5.1
- n(n+1)}{2}(n^2+3 n+1)\end{align} =====Question 9(ii)===== Find the sum of the $n$ terms of the series
- Question 7 Review Exercise
- n^3+8 n^2+6 n-5)}{6}\end{align} =====Question 7(ii)===== Find the sum the series: $3.1^2+5.2^2+7.3^2
- Question 9 Review Exercise
- $S_n=\dfrac{n(2 n^2+6 n+9)}{6}$$ =====Question 9(ii)===== Find the sum of the first $n$ terms of the