Arithmetico - Geometric Series (A.G.S.)

If $\mathrm{a} _{1}, \mathrm{a} _{2}$,………………$a _{n}$ is an A.P. and $b _{1}, b _{2}$,……………..$\mathrm{b} _{\mathrm{n}}$ is a G.P, then the sequence $\mathrm{a} _{1} \mathrm{~b} _{1}, \mathrm{a} _{1} \mathrm{~b} _{2}$,…………….$a _{n} b _{n}$ is said to be an A.G.S. The sequence is of the form $a b,(a+d) b r,(a+2 d) b r^{2}$,……………. ………………

Sum to $n$ terms $=S _{n}=\dfrac{a b}{1-r}+\dfrac{d b r\left(1-r^{n-1}\right)}{(1-r)^{2}}-\dfrac{(a+(n-1) d) b r^{n}}{1-r}$

If $-1<\mathrm{r}<1$, sum to infinite numbers is given by

$\mathrm{S} _{\infty}=\dfrac{\mathrm{ab}}{1-\mathrm{r}}+\dfrac{\mathrm{dbr}}{(1-\mathrm{r})^{2}}$

Important results

1. $\quad$ Let $S _{r}=1^{r}+2^{r}+3^{r}+$…………………$+\mathrm{n}^{\mathrm{r}}$, then

(i) $\mathrm{S} _{1}=1+2+3+$…………………$+\mathrm{n}=\dfrac{\mathrm{n}(\mathrm{n}+1)}{2}$

(ii) $\mathrm{S} _{2}=1^{2}+2^{2}+3^{3}+$…………………$.+\mathrm{n}^{2}=\dfrac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

(iii) $\mathrm{S} _{3}=1^{3}+2^{3}+3^{3}+$…………………$.+\mathrm{n}^{3}=\dfrac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4}=\mathrm{S} _{1}{ }^{2}$

(iv) $\mathrm{S} _{4}=1^{4}+2^{4}+3^{4}+$…………………$+n^{4}=\dfrac{n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)}{30}=\dfrac{S _{2}}{5}\left(6 S _{1}-1\right)$

(iv) $\mathrm{S} _{5}=1^{5}+2^{5}+3^{5}+$…………………$+n^{5}=\dfrac{n^{2}(n+1)^{2}\left(2 n^{2}+2 n-1\right)}{12}=\dfrac{1}{3} S _{1}^{2}\left(4 S _{1}-1\right)$

2. $\quad 1+3+5+$…………………to $\mathrm{n}$ terms $=\mathrm{n}^{2}$

3. $\quad 1^{2}+3^{2}+5^{2}+$…………………to $\mathrm{n}$ terms $=\dfrac{\mathrm{n}\left(4 \mathrm{n}^{2}-1\right)}{3}$

4. $\quad 1^{3}+3^{3}+5^{3}+$…………………to $\mathrm{n}$ terms $=\mathrm{n}^{2}\left(2 \mathrm{n}^{2}-1\right)$

5. $\quad 1-1+1$ - …………………to $\mathrm{n}$ terms $=\dfrac{1-(-1)^{\mathrm{n}}}{2}$

6. $\quad 1-2+3$ -…………………to $\mathrm{n}$ terms $=\dfrac{1-(-1)^{\mathrm{n}}(2 \mathrm{n}+1)}{4}$

7. $\quad 1^{2}-2^{2}+3^{2}$ -…………………to $\mathrm{n}$ terms $=\dfrac{(-1)^{\mathrm{n}-1} \mathrm{n}(\mathrm{n}+1)}{2}=(-1)^{\mathrm{n}-1} \mathrm{~S} _{1}$

8. $\quad 1^{3}-2^{3}+3^{3}$ -…………………to $\mathrm{n}$ terms $=\dfrac{(-1)^{n-1}\left(4 n^{3}+6 n^{2}-1\right)-1}{8}$

Note 1 : $(x+1)(x+2)(x+3)$…………………$(x+n)=x^{n}+A _{1} x^{n-1}+A _{2} x^{n-2}+A _{3} x^{n-3}+$…………………

Then $\quad \mathrm{A} _{1}=\dfrac{\mathrm{n}(\mathrm{n}+1)}{2}$

$\begin{aligned} & \mathrm{A} _{2}=\dfrac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)(3 \mathrm{n}+2)}{24} \\ & \mathrm{~A} _{3}=\dfrac{(\mathrm{n}-1)(\mathrm{n}-2) \mathrm{n}^{2}(\mathrm{n}+1)^{2}}{48} \end{aligned}$

Note 2 : To obtain the sum $\sum\limits _{i<j} a _{i} a _{j}$ we use the identity

$2 \sum\limits _{i<j} a _{i} a _{j}=\left(a _{1}+a _{2}+\ldots \ldots \ldots+a _{n}\right)^{2}-\left(a _{1}{ }^{2}+a _{2}^{2}+\ldots+a _{n}^{2}\right)$

More methods of summation of series

If $n^{\text {th }}$ term of a sequence is given by

$\mathrm{T} _{\mathrm{n}}=\mathrm{an}^{3}+\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{R}$, then

$\begin{aligned} \mathrm{S} _{\mathrm{n}}=\sum \mathrm{T} _{\mathrm{n}} & =\mathrm{T} _{1}+\mathrm{T} _{2}+\ldots \ldots \ldots \ldots \ldots+\mathrm{T} _{\mathrm{n}} \\ & =\mathrm{a} \sum \mathrm{n}^{3}+\mathrm{b} \sum \mathrm{n}^{2}+\mathrm{c} \sum \mathrm{n}+\mathrm{d} \sum 1 \end{aligned}$

I. Method of differences

If the differences of successive terms of a series are in A.P. or G.P., we can find $\mathrm{T} _{\mathrm{n}}$ as follows

(a) Denote $\mathrm{n}^{\text {th }}$ term and the sum up to $\mathrm{n}$ terms by $\mathrm{T} _{\mathrm{n}}$ & $\mathrm{~S} _{\mathrm{n}}$ respectively

(b) Rewrite the given series with each term shifted by one place to the right

(c) Subtracting the above two forms of the series, find $T _{n}$.

(d) Apply $S _{n}=\sum T _{n}$.

Note: Instead of determining the $\mathrm{n}^{\text {th }}$ item of a series by the method of difference, we can use the following steps to obtain the same

(i) If the differences $T _{2}-T _{1}, T _{3}-T _{2}$……………..etc are in A.P. Then take the $\mathrm{n}^{\text {th }}$ term as $\mathrm{T} _{\mathrm{n}}=\mathrm{an}^{2}+\mathrm{bn}+\mathrm{c}, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$

Determine $a, b$, by putting $n=1,2,3$ and equating them with the values of corresponding terms of the given series.

(ii) If the differences $\mathrm{T} _{2}-\mathrm{T} _{1}, \mathrm{~T} _{3}-\mathrm{T} _{2}$………………etc are in G.P, with common ratio $r$, then take

$\mathrm{T} _{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}+\mathrm{bn}+\mathrm{c}, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$

Determine $a, b$, c by putting $n=1,2,3$ and equating them with the values of corresponding terms of the given series.

(iii) If the differences of the differences computed in step (i) are in A.P, then take $T _{n}=a^{3}$ $+\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$

Determine a, $b$, c by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

(iv) If the differences of differences computed in step (i) are in G.P with common ratio r, then take

$T _{n}=a^{n-1}+b^{2}+c n+d$

Determine $a, b$, c by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

II. Sum of series whose $\mathrm{n}^{\text {th }}$ term is

$T _{n}=\dfrac{1}{[a+(n-1) d][(a+n d)]}$

Resolve $T _{n}$ into partial factions, (or express the $\mathrm{N}^{r}$ of $T _{n}$ in terms of factors of $D^{r}$ and simplify), then find $T _{1}, T _{2}$,……………..$\mathrm{T} _{\mathrm{n}}$ and add to get $\mathrm{S} _{\mathrm{n}}$

III. Sum of series in special from

(a) Let the series consists of terms whose $\mathrm{n}^{\text {th }}$ term

$T _{n}=\dfrac{1}{a(a+d)(a+2 d) \ldots \ldots \ldots \ldots . .(a+(n-1) d)}$

To find sum of such a series ( factors of $\mathrm{D}^{\mathrm{r}}$ are in A.P.) as shown above, remove the least factor and multiply the denominator by the number of factors left out (here $\mathrm{n}-1$ ), and also by the common difference (here d) change the sign and add a constant $C$.

Thus, $S _{n}=\dfrac{1}{(n-1) d(a+d)(a+2 d) \ldots \ldots \ldots \ldots \ldots . . .(a+(n-1) d)}+C$

Find $S _{1}$ and hence value of $C$. This gives the required sum.

(b) Let the series consists of terms whose $\mathrm{n}^{\text {th }}$ term $\mathrm{T} _{\mathrm{n}}=\mathrm{a}(\mathrm{a}+\mathrm{d})(\mathrm{a}+2 \mathrm{~d}) \ldots \ldots \ldots . .(\mathrm{a}+(\mathrm{n}-1) \mathrm{d})$.

To find sum of such a series, as shown above, add one more factor and divide by the total number of factors (here ( $\mathrm{n}+1$ ) ) and also by the common difference (here d). Also add a constant $\mathrm{C}$.

Thus $S _{n}=\dfrac{a(a+d)(a+2 d) \ldots \ldots .(a+(n-1) d)(a+n d)}{(n+1) d}+C$

Find $S _{1}$ and hence value of $C$. This gives the required sum.

Note :

(i) for odd $\mathrm{n}, \quad \mathrm{n}=\left(\dfrac{\mathrm{n}+1}{2}\right)^{2}-\left(\dfrac{\mathrm{n}-1}{2}\right)^{2}$

(ii) For any $\mathrm{n}, \quad \mathrm{n}^{3}=\left(\dfrac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}-\left(\dfrac{\mathrm{n}(\mathrm{n}-1)}{2}\right)^{3}$

(iii) For odd $\mathrm{n}, \quad \mathrm{n}^{3}=\left(\dfrac{\mathrm{n}^{3}+1}{2}\right)^{3}-\left(\dfrac{\mathrm{n}^{3}-1}{2}\right)^{2}=\left(\dfrac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}-\left(\dfrac{\mathrm{n}(\mathrm{n}-1)}{2}\right)^{2}$

SOLVED EXAMPLES

1. It is given that $\dfrac{1}{1^{4}}+\dfrac{1}{2^{4}}+\dfrac{1}{3^{4}}+\ldots \ldots \ldots \ldots \infty=\dfrac{\pi^{4}}{90}$ then, $\left \lvert+\dfrac{1}{3^{4}}+\dfrac{1}{5^{4}}+\right.$…………….$\infty$ is equal to

(a) $\dfrac{\pi^{4}}{96}$

(b) $\dfrac{\pi^{4}}{45}$

(c) $\dfrac{89 \pi^{4}}{90}$

(d) None of these

2. If $S _{n}=\sum\limits _{r=1}^{n} \dfrac{1+2+2^{2}+\ldots \ldots \ldots .+r \text { terms }}{2^{r}}$, then

$\mathrm{S} _{\mathrm{n}}$ is equal to

(a) $2^{\mathrm{n}}-(\mathrm{n}+1)$

(b) $1-2^{-n}$

(c) $\mathrm{n}-1+2^{-\mathrm{n}}$

(d) $2^{\mathrm{n}}-1$

3. The sum to $n$ terms of the series

$1^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots=\dfrac{\mathrm{n}(\mathrm{n}+1)^{2}}{2}$ when $\mathrm{n}$ is even. When $\mathrm{n}$ is odd, the sum is

(a) $\dfrac{\mathrm{n}^{2}(\mathrm{n}+1)}{2}$

(b) $\dfrac{\mathrm{n}\left(\mathrm{n}^{2}-1\right)}{2}$

(c) $\mathrm{n}(\mathrm{n}+1)^{2}(2 \mathrm{n}+1)$

(d) None of these

4. If the sum to $\mathrm{n}$ terms of an A.P is $\mathrm{cn}(\mathrm{n}-1) ; \mathrm{c} \neq 0$, then the sum of squares of these terms is

(a) $\mathrm{c}^{2} \mathrm{n}^{2}(\mathrm{n}+1)^{2}$

(b) $\dfrac{2 c^{2}}{3} n(n-1)(2 n-1)$

(c) $\dfrac{2 \mathrm{c}^{2}}{3} \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)$

(d) None of these

5. Let $\mathrm{t} _{\mathrm{r}}=2^{\mathrm{r} / 2}+2^{-\mathrm{r} / 2}$

The $\sum\limits _{\mathrm{r}=1}^{10} \mathrm{t} _{\mathrm{r}}{ }^{2}$ is equal to

(a) $\dfrac{2^{21}-1+20}{2^{10}}$

(b) $\dfrac{2^{21}-1+19}{2^{10}}$

(c) $\dfrac{2^{21}-1-1}{2^{20}}$

(d) None of these

6. Sum to $\mathrm{n}$ terms

1.$(3 n-1)+2$. $(3 n-2)+3 .(3 n-3)+$ .n terms is

(a) $\dfrac{\mathrm{n}(2 \mathrm{n}+1)(5 \mathrm{n}-1)}{3}$

(b) $\dfrac{\mathrm{n}(2 \mathrm{n}+1)(5 \mathrm{n}+1)}{3}$

(c) $\dfrac{\mathrm{n}(\mathrm{n}+1)(7 \mathrm{n}-1)}{6}$

(d) None of these.

7. $\sum\limits _{r=1}^{n} r^{2}-\sum\limits _{m=1}^{n} \sum\limits _{r=1}^{m} r$ is equal to

(a) $\dfrac{1}{2}\left(\sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{2}+\sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}\right)$

(b) $\dfrac{1}{2}\left(\sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{2}-\sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}\right)$

(c) 0

(d) None of these

Exercise

1. For a positive integer $n$ let $\mathrm{a}(\mathrm{n})=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots \ldots \ldots \ldots \ldots+\dfrac{1}{\left(2^{\mathrm{n}}\right)-1}$ then

(a) $\mathrm{a}(100) \leq 100$

(b) $\mathrm{a}(100)>100$

(c) $\mathrm{a}(200) \leq 100$

(d) $\mathrm{a}(200)>100$

2. $11^{3}-10^{3}+9^{3}-8^{3}+7^{3}-6^{3}+5^{3}-4^{3}+3^{3}-2^{3}+1^{3}=$

(a) 756

(b) 724

(c) 648

(d) 812

3. If $a _{1}, a _{2}, \ldots \ldots . . a _{n+1}$ are in A.P with common difference d, then $\sum\limits _{r=1}^{n} \tan ^{-1} \dfrac{d}{1+a _{r} a _{r+1}}$

(a) $\tan ^{-1} \dfrac{n d}{1+a _{1} a _{n+1}}$

(b) $\tan ^{-1} \dfrac{(n+1) d}{1+a _{1} a _{n+1}}$

(c) $\tan ^{-1} \dfrac{(\mathrm{n}-1) \mathrm{d}}{1-\mathrm{a} _{1} \mathrm{a} _{\mathrm{n}+1}}$

(d) $\tan ^{-1} \dfrac{(n-1) d}{1-a _{1} a _{n+1}}$

4. The sum to 50 terms of $\dfrac{3}{1^{2}}+\dfrac{5}{1^{2}+2^{2}}+\dfrac{7}{1^{2}+2^{2}+3^{2}}+$………………..is

(a) $\dfrac{50}{17}$

(b) $\dfrac{100}{17}$

(c) $\dfrac{150}{17}$

(d) $\dfrac{200}{17}$

5. The sum of the first 10 commons terms of the series 17, 21, 25,

(a) 1100

(b) 1010

(c) 1110

(d) 1200

6. Match the following :

7. The sum of the series $\sum\limits _{\mathrm{r}=1}^{\infty} \operatorname{cosec}^{-1} \sqrt{4 \mathrm{r}^{4}+1}$ is

(a) $\pi$

(b) $\pi / 2$

(c) $\pi / 4$

(d) None of these

8. $\quad$ Let $\sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{4}=f(\mathrm{n})$. then $\sum\limits _{\mathrm{r}=1}^{\mathrm{n}}(2 \mathrm{r}-1)^{4}$ is equal to

(a) $f(2 \mathrm{n})-16 f(\mathrm{n}), \forall \mathrm{n} \in \mathrm{N}$

(b) $f(\mathrm{n})-16 f\left(\dfrac{\mathrm{n}-1}{2}\right)$ when $\mathrm{n}$ is odd

(c) $f(\mathrm{n})-16 f\left(\dfrac{\mathrm{n}}{2}\right)$ when $\mathrm{n}$ is odd

(d) None of these

9. Match the following :-

10. If $S=\dfrac{1}{3^{2}+1}++\dfrac{1}{4^{2}+2}+\dfrac{1}{5^{2}+3}+\dfrac{1}{6^{2}+4}+\ldots \ldots \ldots \ldots \infty$ then the value of $\left[\dfrac{1}{S}\right]$ is ……………….

11. The value of the ratio $\left(1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\ldots \ldots \ldots ..\right),\left(1-\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}+\ldots \ldots \ldots \ldots \ldots . ..\right)$ is …………………..

12. $\lim _{\mathrm{n} \rightarrow \infty} \sum\limits _{\mathrm{r}=1}^{\mathrm{n}} \dfrac{\mathrm{r}}{1.3 .5 .7 .9 \ldots \ldots \ldots \ldots \ldots \ldots . .(2 \mathrm{r}+1)}$ is equal to

(a) $\dfrac{1}{3}$

(b) $\dfrac{3}{2}$

(c) $\dfrac{1}{2}$

(d) None of these

13. If $\left(1^{2}-\mathrm{t} _{1}\right)+\left(2^{2}-\mathrm{t} _{2}\right)+\ldots \ldots \ldots \ldots . .+\left(\mathrm{n}^{2}-\mathrm{t} _{\mathrm{n}}\right)=\dfrac{\mathrm{n}}{3}\left(\mathrm{n}^{2}-1\right)$, then $\mathrm{t} _{\mathrm{n}}$ is equal to

(a) $\mathrm{n}^{2}$

(b) $2 \mathrm{n}$

(c) $n^{2}-2 n$

(d) None of these

14. If $(1+3+5+$………………$+\mathrm{p})+(1+3+5+$………………$+q)=(1+3+5+$………………$+r$ ) where each set of parantheses contains the sum of consecutive odd integers as shown, the smallest possible value of $\mathrm{p}+\mathrm{q}+\mathrm{r}($ where $\mathrm{p}>6)$ is………………

(a) 12

(b) 21

(c) 45

(d) 54