Calculate Sn (x) = x + 2x^2 + 3x^3 +...+nx^n if x is real and n is natural.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll calculate Sn (x) for x = 1.

Sn (x) = 1 + 2 + 3 + .... + n

We recognize the sum of the first natural n terms:

Sn (1) = n(n+1)/2

Now, we'll multiply Sn (x) by x:

xSn (x) = x^2 + 2x^3 + 3x^4 + ... + (n-1)x^n + nx^(n+1)

We'll calculate the difference:

Sn (x) - xSn (x) = x + x^2 + x^3 + ...+x^n - nx^(n+1)

We'll recognize the sum of the terms of the geometric progression, whose common ratio is x:

x + x^2 + x^3 + ...+x^n = [x - x^(n+1)]/(1-x)

Sn (x) - xSn (x) = [x - x^(n+1)]/(1-x) - nx^(n+1)

Sn (x) - xSn (x) = [x - (n+1)x^(n+1) + nx^(n+2)]/(1-x)

So, for x = 1, Sn(x) = n(n+1)/2

For any value of x, except 1:

Sn(x) = [nx^(n+2) - (n+1)x^(n+1) + x]/(1-x)

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neela | High School Teacher | (Level 3) Valedictorian

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To calculate Sn(x) = x+2x^2+3x^3 +...nx^n.

Solution:

Sn (x)  =  x +2x^2+3x^3+4x^4 +.......nx^n...........(1)

xSn(x) =         x^2 +2x^3 +3x^2 ...... (n-1)x^n + nx^(n+1)....(2)

Subtracting we get:

(1-x)Sn(x) = (x+x^2+x^3 +.....nx^n )+ nx^(n+1).

(1-x)Sn(x) = x(1+x+x^2+....x^n-1) + nx^(n+1).

(1-x)Sn(x) = x(1-x^n)/(1-x) + nx^(n+1).

Sn(x) = x(1-x^n)/(1-x)^2 + nx^(n+1)/(1-x).

Or

Sn(x) = x(x^n  -1)/(x-1)^2 +nx^(n+1)/(x-1)

Therefore the the sum of the series,

x+2x^2 +3x^3 +4x^4+...nx^n = x(x^n  -1)/(x-1)^2 + x^(n+1)/(x-1).

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