Sequence and Series

1. $\frac{1}{2},-\frac{2}{3},\frac{3}{4},-\frac{4}{5}...$ Ans: $(-1{)}^{n+1}\cdot \frac{n}{n+1}$ 

        $1\cdot \frac{1}{1+1},-1\cdot \frac{2}{2+1},1\cdot \frac{3}{3+1},-1\cdot \frac{4}{4+1}\dots$ 

      

2. $\frac{1}{2},\frac{3}{4},\frac{5}{6},\frac{7}{8}...$ Ans: $\frac{2n-1}{2n}$ 

3. $\frac{1}{2},\frac{1}{2},\frac{3}{2^3},\frac{4}{2^4},...$ Ans: $\frac{n}{2^n}$ 

4. $1,\sqrt{2},\sqrt{3},2,...$ Ans: $\sqrt{n}$ 

n^th term = ar^n-1 

1. Sum of the three consecutive terms of a geometric series is 21/2 and product is 8. Find the 3^rd term. 

solution: 

a + ar + ar² = 21/2 ............(i)    and    a.ar.ar² = 8 ..........(ii)

From (ii), (ar)³ = 8 or, a = 2/r .......... (iii)

Substitute (iii) into (i), r=4 or r=1/4

when r=4, a=1/2 , 3rd term is 8

when r=1/4 a=8 , 3rd term is 1/2

$$S_n=\frac{a(r^n-1)}{r-1}\text{when r>1}$$ $$S_n=\frac{a(1-r^n)}{1-r}\text{when r<1}$$* Actually either of the formulas can be used without taking the value of r into consideration. We use different versions of the same formula based on the value of r to avoid negative number in denominator. 

1. 4 + 44 + 444 + ... ... ... 

Solution: 

S_n = 4 + 44 + 444 + ... + n^th term

    = 4 (1 + 11 + 111 + ... + nth term)

    = $\frac{4}{9}$ (9 + 99 + 999 + ... + nth term)

    = $\frac{4}{9}$ { (10-1) + (100-1) + (1000-1) + ... + nth term }

If the absolute value of the common ratio r of a geometric series is less than 1 then sum to infinity of the series exists. So, the condition for a geometric series to have sum to infinity is:

$$|r|<1$$ 
$$\text{or,}-1<r<1$$ 

$$S_{\mathrm{\infty }}=\frac{a}{1-r}$$ 

1. $5+\frac{10}{3}+\frac{20}{9}+\frac{40}{27}+...$ [Dhaka Board 2023]

2. 

1. 2(3x - 5)^-1 + 4(3x - 5)^-2 + 8(3x - 5)^-3 + ... ... ... [Dhaka Board 2024]

2. (3x + 1)-1 + (3x + 1)-2 + (3x + 1)-3 + ... ... ... [Cumilla Board 2024]

3. (2x + 3)-1 + (2x + 3)-2 + (2x + 3)-3 + ... ... ... [Mymensigh Board 2024]

4. 1/(5x-4)^-1 + 1/(5x-4)^-2 + 1/(5x-4)^-3 + ... ... ... [Barishal Board 2022] Solution: x<3/5 or x>1 sum to infinity = 1/5(x-1)