\[\large{\displaystyle \sum_{n=1}^{P} \frac{n^2}{2^n} + \displaystyle \sum_{n=1}^{P} \frac{n}{2^n}}\]

how am I supposed to express this as an equation?......plz help! :( :(

\[\large{\displaystyle \sum_{n=1}^{P} \frac{n^2}{2^n} + \displaystyle \sum_{n=1}^{P} \frac{n}{2^n}}\]

how am I supposed to express this as an equation?......plz help! :( :(

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestMy solution uses differential calculus. We start with the formula for an infinite geometric series: \[\sum_{n=1}^{\infty} x^n = \dfrac{x}{1-x} = \dfrac{1}{1-x} - 1\] Now if we differentiate this with respect to \(x\), we get: \[\sum_{n=1}^{\infty} n x^{n-1} = \dfrac{1}{(1-x)^2} \to (1)\] Differentiating one more time gives us: \[\sum_{n=1}^{\infty} n(n-1)x^{n-2} = \dfrac{2}{(1-x)^3} \to (2)\]

Multiplying \((1)\) by \(2x\) gives us: \[\sum_{n=1}^{\infty} 2n x^n = \dfrac{2x}{(1-x)^2} \to (3)\]

Multiplying \((2)\) by \(x^2\) gives us: \[\sum_{n=1}^{\infty} n(n-1)x^n = \dfrac{2x^2}{(1-x)^3} \to (4)\]

Now after adding \((3)\) and \((4)\) we get: \[\begin{array}{ccl}\displaystyle\sum_{n=1}^{\infty} (n^2+n) x^n & = & \dfrac{2x^2}{(1-x)^3} + \dfrac{2x}{(1-x)^2} \\ \displaystyle\sum_{n=1}^{\infty} (n^2+n) x^n & = & \dfrac{2x}{(1-x)^3} \to (5) \end{array}\]

Of course, this is an infinite series. Let's calculate the formula we need using the formulas above:

\[\begin{array}{ccl} \sum_{n=1}^{P} (n^2+n)x^n & = & \sum_{n=1}^{\infty} (n^2+n)x^n - \sum_{n=P+1}^{\infty} (n^2+n)x^n \\ & = & \dfrac{2x}{(1-x)^3} - \sum_{n=1}^{\infty} \left((n+P)^2+(n+P)\right)x^{n+P} \\ & = & \dfrac{2x}{(1-x)^3} - x^P \left[\sum_{n=1}^{\infty} (n^2+n)x^n + P \sum_{n=1}^{\infty} 2nx^n + (P^2+P) \sum_{n=1}^{\infty} x^n \right] \\ & = & \dfrac{2x}{(1-x)^3} - x^P \left[\dfrac{2x}{(1-x)^3} + \dfrac{2Px}{(1-x)^2} + \dfrac{(P^2+P)x}{1-x} \right] \end{array}\]

Therefore, substituting \(x = \dfrac{1}{2}\) gives us:

\[\sum_{n=1}^{P} \dfrac{n^2+n}{2^n} = 8 - \dfrac{8 + 5P + P^2}{2^P} \] – Ariel Gershon · 7 months, 1 week ago

Log in to reply

\[\large{\displaystyle \sum_{n=1}^{\infty} \frac{n}{2^n + 3^n}}\] – Asif Hasan · 6 months, 2 weeks ago

Log in to reply

– Ariel Gershon · 6 months, 2 weeks ago

You're welcome! Oh wow, I don't know... There might not even be a closed form for that sum.Log in to reply

– Asif Hasan · 6 months, 2 weeks ago

ohh :( :( thanks for the reply anyway!!Log in to reply

– Asif Hasan · 6 months, 2 weeks ago

Thanks a lot!! Very elegant solution! :D thanks!!Log in to reply

\[\large{\displaystyle \sum_{n=1}^{P} \frac{n^2}{2^n} + \displaystyle \sum_{n=1}^{P} \frac{n}{2^n}}\]

Hit on your profile picture at the right top corner of the page, then hit "Toggle Latex". Copy the latex code and then do whatever you want to :P – Satyajit Mohanty · 1 year, 10 months ago

Log in to reply

– Asif Hasan · 1 year, 10 months ago

done.......:) :)Log in to reply