Question

1 × 3^1 + 2 × 3^2 + 3 × 3^3 + … + n × 3^n = 3/4 (35*3^18 + 1)

What is the value of n?

Answer 1

I cant think of a short way to do this problem.
I have a pretty lengthy solution that shows n = 18

Answer 2

This is the sum of the terms of a sequence of the form
\[U(n)=aq^n\]
With a=n and q=3.
The sum n terms is given by
\[s_{n}=a \frac{1-q^n}{1-q}\]
I'd try starting with that no?

Answer 3

The way I did it was:
First use the geometric series formula:
\[1+x+x^2+\cdots+x^n = \frac{x^{n+1}-1}{x-1}\] Take the derivative of both sides:
\[1+2x+3x^2+\cdots+nx^{n-1} = \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}\]Multiply both sides by x:
\[x+2x^2+3x^3+\cdots+nx^n = \frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}\]
Notice if you let x = 3, the left hand side becomes the same as the left hand side of the problem that was asked. So plug x = 3 into the right hand side, and simplify, then set equal to the right hand side of the problem. You will end up with n = 18.
Like i said, i dont know if there is a shorter way.

Answer 4

When x = 3 you get:
\[\frac{n(3)^{n+2}-(n+1)3^{n+1}+3}{(3-1)^2} = \frac{9n(3)^n-3(n+1)(3)^n+3}{4} \]
\[\frac{(9n-3n-3)(3)^n+3}{4} = \frac{(6n-3)3^n+3}{4} = \frac{3}{4}((2n-1)3^n+1) \]
Setting that equal to the right hand side of the problem we obtain:
\[\frac{3}{4}((2n-1)3^n+1) = \frac{3}{4}(35(3)^n+1)\]
Thus we have:
\[2n-1 = 35, n = 18\]

Answer 5

very nice..i never thought of taking the derivative.
well the easy way of solving this is using a program or excel and compute it numerically since we know what the sum is.
however, mathematically speaking
i found it can be broken down into a series of smaller geometric series
\[3^{1} + (2)3^{2}...+(n)3^{n} = (3^{1}+3^{2}...+3^{n}) + (3^{2}+3^{3}...+3^{n})...+3^{n}\]
using formula for geometric sum for each one
\[\frac{(3^{n+1}-3)+(3^{n+1}-3^{2})...+(3^{n+1}-3^{n})}{3-1}\]
which can reduce to
\[\frac{3[3^{n}(2n-1)]}{4} \]

Answer 6

wow that is also very nice, im going to take not of that :)