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 $\mathrm{a} _{1} \mathrm{~b} _{1}, \mathrm{a} _{1} \mathrm{~b} _{2}$, $…………\mathrm{a} _{\mathrm{n}} \mathrm{b} _{\mathrm{n}}$ 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 $=S _{n}=\frac{a b}{1-r}+\frac{d b r\left(1-r^{n-1}\right)}{(1-r)^{2}}-\frac{(a+(n-1) d) b r^{n}}{1-r}$

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

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

Important results

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

$\quad$ (i) $\mathrm{S} _{1}=1+2+3+\ldots \ldots \ldots \ldots \ldots+n=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

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

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

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

$\quad$ (iv) $\mathrm{S} _{5}=1^{5}+2^{5}+3^{5}+\ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{n}^{5}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}\left(2 \mathrm{n}^{2}+2 \mathrm{n}-1\right)}{12}=\frac{1}{3} \mathrm{~S} _{1}^{2}\left(4 \mathrm{~S} _{1}-1\right)$

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

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

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

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

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

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

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

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

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

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

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

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

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 $\mathrm{T} _{2}-\mathrm{T} _{1}, \mathrm{~T} _{3}-\mathrm{T} _{2},……….$ 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 $a, b$, 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 $\mathrm{a}, \mathrm{b}$, c by putting $\mathrm{n}=1,2,3,4$ 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

$\mathrm{T} _{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}+\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$

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

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

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

Resolve $T _{n}$ into partial fractions, (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 form

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

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

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

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

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

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

Note :

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

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

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

Solved examples

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

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

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

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

(d) None of these

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

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

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

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

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

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

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

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

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

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

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

(d) None of these

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

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

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

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

(d) None of these

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

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

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

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

(c) $\frac{2^{21}-1}{2^{20}}-19$

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

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

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

(d) None of these.

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

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

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

(c) $0$

(d) None of these

Practice questions

1. For a positive integer $n$ let

$\mathrm{a}(\mathrm{n})=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\left(2^{\mathrm{n}}\right)-1}$ then

(a) $a(100) \leq 100$

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

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

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

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

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

4. The sum to 50 terms of

$\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+……….$ is

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

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

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

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

5. The sum of the first 10 common terms of the series $17, 21, 25,………$ and $16,21,25, ……..$ is

(a) 1100

(b) 1010

(c) 1110

(d) 1200

6. Match the following:

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

(a) $\pi$

(b) $\pi / 2$

(c) $\pi / 4$

(d) None of these

8. Let $\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{4}=f(\mathrm{n})$, then $\sum _{\mathrm{r}=1}^{\mathrm{n}}(2 \mathrm{r}-1)^{4}$ 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(\frac{\mathrm{n}-1}{2}\right)$ when $\mathrm{n}$ is odd

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

(d) None of these

9. Match the following:-

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

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

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

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

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

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

(d) None of these

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

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

(b) $2 \mathrm{n}$

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

(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