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- Question 1 Review Exercise 5
- the series: $1+\dfrac{2}{3}+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\ldots$ * %%(a)%% $6$ ... ollapse> iii. Sum the series:$1+2.2+3.2^2+\cdots+100.2^{\prime \prime}$ * (a) $99.2^{100}$ * (b) $100.2^{100}$ * %%(c)%% $99.2^{100}+1$ * %%(d)%% $1000.2^{100}$ \\ <btn type="
- Question 4 Review Exercise
- Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product of the succ... dfrac{1}{7})+(\dfrac{1}{4}-\dfrac{2}{7}+\dfrac{1}{10})+ \\ &\cdots+(\dfrac{1}{3 n-2}-\dfrac{2}{3 n+1}+... rac{1}{18}[(1+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{10} \cdots+\dfrac{1}{3 n-2}) \\ & -(\dfrac{2}{4}+\df
- Question 3 Exercise 5.3
- $ term and sum to $n$ terms each of the series $4+10+18+28+40+\ldots$ ====Solution==== We use the method of difference as: \begin{align} & a_2-a_1=10-4=6 \\ & a_3-a_2=18-10=8 \\ & a_4-a_3=28-18=10 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n \quad 1}=(\
- Question 2 & 3 Exercise 5.1
- Pakistan. Q2 Find the sum $1.2+2.3+3.4+\ldots+99.100$. Solution: The given series is the product of t... of the series $1+2+3+\ldots+99$ and $2+3+4+\ldots+100$, whose $n^{\text {th }}$ terms are $n(n+1)$ and... \end{aligned} $$ $$ \begin{aligned} & =\frac{99(100)(199)}{6}+\frac{99(100)}{2} \\ & =(33.50 .199)+(99.50) \\ & \Rightarrow 1.2+2.3+3.4+\ldots+99.100 \\
- Question 1 Exercise 5.2
- ===== Sum up to $n$ terms the series $1+4 x+7 x^2+10 x^3+\ldots.$ ====Solution==== Let \begin{align} & S_n=1+4 x+7 x^2+10 x^3+\ldots +(3 n-2) x^{n-1}....(i) \\ & x S_n=x+4 x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align} Su... et \begin{align} (1-x) S_n&=1+(4-1) x+(7-4) x^2+ (10-7) x^3+\ldots \\ & +(3 n-2,3 n-5)) x n-1-(3 n-2)
- Question 2 Exercise 5.3
- ts$ ====Solution==== \begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16 \\ & a_4-a_3=52-30=22 \\ & ... _{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,16,22, \ldots\end{align} which is a A.P. Adding column wise, we get \begin{align} a_n-a_1&=10+16+22+\ldots+(n-1) \text { terms } \\ & =\dfrac{n-1}{2}[2 \cdot 10+(n-2) \cdot 6] \\ & =\dfrac{n-1}{2}[20+6 n-12] \\
- Question 1 Exercise 5.1
- n}& \therefore T_j-(4 j-3)^3 \\ & =64 j^3-144 j^2+108 j-27\end{align} Taking sum of the both sides of ... j=64 \sum_{j=1}^n j^3-144 \sum_{j=1}^n j^2 \\ & +108 \sum_{j=1}^n j-27 \sum_{j=1}^n 1 \\ & =64(\dfrac{n(n+1)}{2})^2-144 \dfrac{n(n+1)(2 n+1)}{6}+108 \dfrac{n(n-1)}{2}-27 n \\ & =64 \dfrac{n^2(n+1)^
- Question 6 Exercise 5.3
- 32-28=4 \\ & a_3-a_2=52-32=20 \\ & a_4-a_3=152-52=100 \\ & \ldots \quad \cdots \quad \cdots \\ & \cdot... (\mathrm{n}-1) \text { term ofthe sequence } 4,20,100, \ldots \end{align} which is a G.P with common r... g column wise, we get \begin{align} a_n-a_1&=4+20+100+\ldots+(n-1) \text { terms } \\ & =\dfrac{4[5^{n
- Question 8 Review Exercise
- =\dfrac{n(n+1)}{2} \cdot[\dfrac{n(n+1)}{2}+\dfrac{10 n+5}{3}+4] \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{3 n(n+1)+2(10 n+5)+24}{6}] \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{3 n^2+3 n+20 n+10+24}{6}] \\ & =\dfrac{n(n+1)(3 n^2+23 n+34)}{12}\e
- Question 10 Review Exercise
- ====== Question 10 Review Exercise ====== Solutions of Question 10 of Review Exercise of Unit 05: Miscullaneous Series. ... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10===== Find the $n^{\text {th }}$ term and the sum
- Question 4 & 5 Exercise 5.1
- }(n+1)\end{align} =====Question 5===== Sum: $2+5+10+17+\ldots$ to $n$ terms. ====Solution==== First w
- Question 7 & 8 Exercise 5.1
- ===Question 7===== Sum to $n$ terms: $1.5 .9+2.6 .10+3.7 .11+\ldots$ ====Solution==== The general term
- Question 4 Exercise 5.3
- n}-1) \text { term ofthe sequence }\end{align} $6,10,18, \ldots$ which is a G.P. Adding column wise, w
- Question 9 Review Exercise
- lign="right"><btn type="success">[[math-11-kpk:sol:unit05:Re-ex5-p8|Question 10 >]]</btn></text>