Sum of simple series

Question

What is the summation of this series ?

$1 \times \frac{1}{2}$ + $2 \times \frac{1}{4}$ +$3 \times \frac{1}{8}$ + .....

Answer 1

An arithmetico-geometric sequence[A.G.P] is the result of the multiplication of a geometric progression with the corresponding terms of an arithmetic progression

S(n)=1*$\frac{1}{2}$+2*$\frac{1}{4}$+3*$\frac{1}{8}$+4*$\frac{1}{16}$+............Equation(1)

Multiply both sides by common ratio(r) of gp series =$\frac{1}{2}$

$\frac{1}{2}$S(n)=1*$\frac{1}{4}$+2*$\frac{1}{8}$+3*$\frac{1}{16}$+........Equation(2)

subtract eq2 from eq1

S(n)-$\frac{1}{2}$S(n)=1*$\frac{1}{2}$+(2-1)*$\frac{1}{4}$+(3-2)$\frac{1}{8}$+(4-3)$\frac{1}{16}$+......

$\frac{1}{2}$S(n)=$\frac{1}{2}$+$\frac{1}{4}$+$\frac{1}{8}$+$\frac{1}{16}$+.....

$\frac{1}{2}$S(n)=$\frac{\frac{1}{2}}{1-(\frac{1}{2})}$

S(n)=2

Answer 2

Let us revise an old formula -> 1 + a + a^2 + a^3 + . . = 1/(1-a) thats our geometric series  
 
1 + 2a + 3a^2 + 4a^3 + . . = (1/(1 - a))' = 1/(1-a)², multiply by a

a + 2a^2 + 3a^3 + . . = a/(1-a)^2, let a = 1/2

1*(1/2) + 2*(1/2)^2 + 3*(1/2)&3 + . . = (1/2)/(1-1/2)^2 = 1/(1/2) = 2

This will be implied even on our logest series . No matter how long we go on calculation... Our answer will converge on 2 .

Kindly correct me if I am wrong

Thank you :)

Comments

good one

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in ur first step ,
1 + 2a + 3a^2 + 4a^3 + . . = (1/(1 - a))' = 1/(1-a)²

How did you get (1/(1 - a))' = 1/(1-a)² ??

Sorry, need more explanations pls..