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- Question 10 Review Exercise
- frac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8})+\ldots$ ====Solution==== The general term of the given series is: \beg
- Question 9 Review Exercise
- $n$ terms of the series $3+7+13+21+31+\ldots$ ====Solution==== Using method of differences to compute the su... st $n$ terms of the series $2+5+14+41+\ldots$ ====Solution==== Using method of differences to compute the su
- Question 8 Review Exercise
- the series whose $n^{t h}$ term is $n^3+3^n.$ ====Solution==== The $n^h$ term is: $$a_n=n^3+3^n$$ Taking sum... he series whose $n^{t h}$ term is $2 n^2+3 n$ ====Solution==== The $n^{t h}$ term is: $$a_n=2 n^2+3 n$$ Tak... series whose $n^{t h}$ term is $n(n+1)(n+4)$ ====Solution==== The $n^{\text {th }}$ term is: \begin{align} ... he series whose $n^{t h}$ term is $(2 n-1)^2$ ====Solution==== The $n^{t h}$ term is: \begin{align} & a_n=(2
- Question 7 Review Exercise
- ies: $1.2^2+3.3^2+5.4^2+\ldots$ to $n$ terms. ====Solution==== The given series if the product of correspond... ies: $3.1^2+5.2^2+7.3^2+\ldots$ to $n$ terms. ====Solution==== In the given series each term is the product
- Question 4 Review Exercise
- dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product o
- Question 5 & 6 Review Exercise
- : $5+12 x+19 x^2+26 x^3+\ldots$ to $n$ terms. ====Solution==== Let \begin{align}S_n&=5+12 x+19 x^2+26 x^3+\c... {1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== Solution: The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fra
- Question 2 & 3 Review Exercise
- the series to $n$ terms $1.2+2.3+3.4+\ldots$ ====Solution==== The $n^{\text {th }}$ term is: $$a_n=n(n+1)=n... : $1.3 .5+2.4 .6+3.5 .7+\ldots$ to $n$ terms. ====Solution==== In the given series each term is the product
- Question 4 Exercise 5.4
- series: $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$ ====Solution==== Let \begin{align}S_n &=\sum_{k=1}^n \dfrac{1}
- Question 2 & 3 Exercise 5.4
- series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n ... f the series: $\sum_{k=1}^n \dfrac{1}{k^2-k}$ ====Solution==== Let $$S_n=\sum_{k=1}^n \dfrac{1}{k^2-k}=\sum_
- Question 6 Exercise 5.3
- each of the series $28+32+52+152+652+\ldots$ ====Solution==== \begin{align} & a_2-a_1=32-28=4 \\ & a_3-a_2=
- Question 1 Exercise 5.3
- {1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== The general term of the series is: $$T_n=\dfr... {1}{3.5}+\dfrac{1}{5.7}+\ldots$ to $n$ terms. ====Solution==== Here $n$ term of the series is: $u_n=\dfrac{1... {1}{5.8}+\dfrac{1}{8.11}+\ldots$ to infinity. ====Solution==== Here in the denominator the factors are the p... {13.22}+\dfrac{1}{22.31}+\ldots$ to infinity. ====Solution==== Here in the deneminator the factors are the p
- Question 5 Exercise 5.3
- each of the series $3+9+21+45+93+189+\ldots$ ====Solution==== \begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21
- Question 4 Exercise 5.3
- each of the series $3+5+11+29+83+245+\ldots$ ====Solution==== \begin{align} & a_2-a_1=5-3=2 \\ & a_3-a_2=11
- Question 2 Exercise 5.3
- each of the series $4+14+30+52+80+114+\ldots$ ====Solution==== \begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=
- Question 3 Exercise 5.3
- rms each of the series $4+10+18+28+40+\ldots$ ====Solution==== We use the method of difference as: \begin{al