What is the formula for the summation of the folllowing series: x*(y) + 2x*(y^2) + 3x*(y^3).......nx*(y^n)?

Question

Please note that I need a formula to calculate the summation to nth term (Sn)

x*(y) + 2x*(y^2) + 3x*(y^3) ....... nx*(y^n)

Answer 1

I did one similar to this for someone yesterday!

x is a common factor, so you just need to sum
y + 2y^2 + 3y^3 + ... + ny^n, and multiply the result by x.

Note that
y + y^2 + y^3 + ... + y^n
= y(y^n - 1)/(y - 1) [geometric series with a = y, r = y]

Differentiate both sides with respect to y, using the quotient formula of course for the right hand side:

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

The numerator of the right hand side reduces to
ny^(n+1) - (n+1)y^n + 1 I think.

Now multiply both sides by xy and get

xy + 2xy^2 + 3xy^3 + nxy^n
= xy(ny^(n+1) - (n+1)y^n + 1)/(y - 1)^2

Answer 2

Let A = xy + 2xy^2 + 3xy^3 + ... + nxy^n
Take out the factor xy to give:
A = xy[1 + 2y + 3y^2 + ... + ny^(n-1)]

Now let B = 1 + 2y + 3y^2 + ... + ny^(n-1), so that A = xyB.
Multiply through by y to give:
By = y + 2y^2 + 3y^3 + ... + ny^n
Now subtract B from By to give:
By - B = -1 - y - y^2 - y^3 - ... - y^(n-1) + ny^n
or, B(y-1) = ny^n - [1 + y + y^2 + y^3 + ... + y^(n-1)]

Now let C = 1 + y + y^2 + y^3 + ... + y^(n-1)
Multiply through by y to give:
Cy = y + y^2 + y^3 + y^4 + ... + y^n
Now subtract C from Cy to give:
Cy - C = -1 + y^n
or, C = (y^n-1) / (y-1)

Substituting back into the previous paragraph gives:
B(y-1) = ny^n - (y^n-1) / (y-1)
Therefore, B = [ny^(n+1) - (n+1)y^n + 1] / (y-1)^2

But A = xyB.

Therefore, A = xy[ny^(n+1) - (n+1)y^n + 1] / (y-1)^2
This is the formula for the summation of n terms.

Answer 3

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Answer 4

Let
s = xy + 2xy² + 3xy³ + ... + nxyⁿ

Factor out x, then divide both sides by x
s/x = y + 2y² + 3y³ + ... + nyⁿ

Now,
sy/x = y² + 2y³ + ... + ny^(n + 1)

We subtract
s/x - sy/x = y + y² + y³ + ... + yⁿ - ny^(n + 1)

On the left side we factor out s/x
s/x (1 - y) = y + y² + y³ + ... + yⁿ - ny^(n + 1)

Since y + y² + y³ + ... + yⁿ is a geometric progression of ratio y, then the sum is
s/x (1 - y) = (y)(yⁿ - 1)/(y - 1) - ny^(n + 1)

We factor out y from the right side
s/x (1 - y) = y[(yⁿ - 1)/(y - 1) - nyⁿ]

We multiply x/(1 - y) to both sides and get the lowest term on the right side
s = xy[(yⁿ - 1 - ny^(n + 1) + nyⁿ]/(1 - y)

Some more simplification
s = xy[(n + 1)yⁿ - ny^(n + 1) - 1]/(1 - y)

Multiply -1/-1 to the right side
s = xy[ny^(n + 1) - (n + 1)yⁿ + 1]/(y - 1)

Therefore, that is the sum!
^_^
^_^

Answer 5

Consider,
1+y+y^2+...+y^n = ( y^(n+1) - 1 )/ (y-1)
Differentiating both sides with respect to y we get,
1+2*y+...+n*y^(n-1)
= (((n+1)(y^n)(y-1) - y^(n+1)+1))/(y-1)^2
= (( (n+1)y^(n+1) - (n+1)y^n - y^(n+1) +1)/(y-1)^2
= ( ny^(n+1) - (n+1)y^n +1)/ (y-1)^2

Now multiplying both sides with x*y we get,
x*y + 2x*(y^2) +...+nx*(y^n)
= x*y*(ny^(n+1) - (n+1)y^n +1)/(y-1)^2

And so we get the answer for the given series!

Answer 6

First factor out x
To treat : B= y + 2y^2 + 3y^3 ...+ny^n
Factor out y
B= y* (1 + 2y + 3y^2 + +ny^(n-1))= y*C
Observe that C is the derivative of
y + y^2 + y^n whose sum is
y* (y^n -1)/(y-1)
You have to find the primitive of this....

I suppose you are now able to continue...

Answer 7

Following the lead of those who answered previously:
xy(d{y + y^2 + y^3 + ... + y^n
= y(y^n - 1)/(y - 1) = (y^(n+1) - y)/(y-1)}/dy =
xy(1+2y+3y^2+4y^3 ...+ny^(n-1)) =
xy[((y-1)((n+1)y^n-1)-(yy^n-y))/(y-1)^2] =
xy[((y-1)(ny^n+y^n-1)-yy^n+y)/(y-1)^2] =
S(x,y) = xy[((y^n)(n(y - 1) - 1) + 1)/(y - 1)^2], y ≠ 1
S(x,1) = xn(n+1)/2