# Help with summation

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

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

Note by Asif Hasan
3 years ago

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My 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:

$\sum_{n=1}^{P} \dfrac{n^2+n}{2^n} = 8 - \dfrac{8 + 5P + P^2}{2^P}$

- 1 year, 9 months ago

hey, a little more help plz!! how are we supposed to handle this type::

$\large{\displaystyle \sum_{n=1}^{\infty} \frac{n}{2^n + 3^n}}$

- 1 year, 8 months ago

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

- 1 year, 8 months ago

ohh :( :( thanks for the reply anyway!!

- 1 year, 8 months ago

Thanks a lot!! Very elegant solution! :D thanks!!

- 1 year, 8 months ago

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

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

- 3 years ago

done.......:) :)

- 3 years ago