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- Question 1 Exercise 5.1
- ies $1^2+3^2+5^2+7^2+\ldots$ up to $n$ terms. ====Solution==== We see that each term of the given series is ... 2+2^2)+(1^2+2^2+3^2)+\ldots$ up to $n$ terms. ====Solution==== In the given series, we see that $T_1=1^2, T_... ies $2^2+4^2+6^2+8^2+\ldots$ up to $n$ terms. ====Solution==== The $n^{t h}$-term of the series is: $T_j=(\d... p to $n$ terms. (iv) $1^3+3^3+5^3+7^3+\ldots$ ====Solution==== The $n^{t h}$-term of the series is: $$T_1=(2
- Question 1 Exercise 5.2
- ms the series $1.2+2.2^2+3.2^3+4.2^4+\ldots$. ====Solution==== Let \begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3... terms the series $1+4 x+7 x^2+10 x^3+\ldots.$ ====Solution==== Let \begin{align} & S_n=1+4 x+7 x^2+10 x^3+\l... terms the series $1+2 x+3 x^2+4 x^3+\ldots$. ====Solution==== Let \begin{align} & S_n=1+2 x+3 x^2+4 x^3+\ld... dfrac{3}{2}+\dfrac{5}{4}+\dfrac{7}{8}+\ldots$ ====Solution==== Let \begin{align} & S_n=1+\dfrac{3}{2}+\dfrac
- Question 1 Exercise 5.3
- {1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== The general term of the series is: $$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 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 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 Exercise 5.1
- an. Q2 Find the sum $1.2+2.3+3.4+\ldots+99.100$. Solution: The given series is the product of the correspon... d} $$ Q3 Find the sum $1^2+3^2+5^2+\ldots+99^2$. Solution: The each term of the given series is the square
- Question 4 & 5 Exercise 5.1
- the $2+(2+5)+(2+5+8)+\ldots$ up to $n$ terms. ====Solution==== The general term of the sequence is: \begin{a... 5===== Sum: $2+5+10+17+\ldots$ to $n$ terms. ====Solution==== First we reform the given series as: $$(1+1^2
- Question 7 & 8 Exercise 5.1
- to $n$ terms: $1.5 .9+2.6 .10+3.7 .11+\ldots$ ====Solution==== The general term of the series is: $T_j=j(j+4... terms, whose $n^{t h}$-term is $4 n^2+5 n+1$. ====Solution==== Taking summation of the general term of the s
- Question 9 Exercise 5.1
- he series whose $n$-term is (32 n^2+54 n+25). ====Solution==== The $n$-term of the the series is given as: \... se $n$-term is $3\left(4^n+2 n\right)-4 n^3$. ====Solution==== The general term of the series is: $$T_j=3(4^
- Question 2 & 3 Exercise 5.2
- 2===== $1+3^2 x+5^2 x^2+7^2 x^3+\ldots, x<1$. ====Solution==== Let \begin{align} & S_{\infty}=1+3^2 x+5^2 x^... rac{3}{8}+\dfrac{4}{16}+\dfrac{5}{32}+\ldots$ ====Solution==== The given series is the product of the corres
- Question 4 & 5 Exercise 5.2
- {7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots$ ====Solution==== Let \begin{align} & S_{\infty}=5+\dfrac{7}{3}... y$ is $\dfrac{44}{9}$, find the value of $r$. ====Solution==== Let $$S_{\infty}=3+5 r+7 r^2+\ldots \infty...
- 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 $$S_n=\sum_{k=1}^n \dfrac{1}{k^2-k}=\sum_
- Question 2 & 3 Review Exercise
- the series to $n$ terms $1.2+2.3+3.4+\ldots$ ====Solution==== The $n^{\text {th }}$ term is: $$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 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 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