Help with summation

∑n=1Pn22n+∑n=1Pn2n\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! :( :(

#Algebra #NumberTheory #Summation

Note by Asif Hasan
5 years, 11 months ago

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Comments

My solution uses differential calculus. We start with the formula for an infinite geometric series: ∑n=1∞xn=x1−x=11−x−1\sum_{n=1}^{\infty} x^n = \dfrac{x}{1-x} = \dfrac{1}{1-x} - 1 Now if we differentiate this with respect to xx, we get: ∑n=1∞nxn−1=1(1−x)2→(1)\sum_{n=1}^{\infty} n x^{n-1} = \dfrac{1}{(1-x)^2} \to (1) Differentiating one more time gives us: ∑n=1∞n(n−1)xn−2=2(1−x)3→(2)\sum_{n=1}^{\infty} n(n-1)x^{n-2} = \dfrac{2}{(1-x)^3} \to (2)

Multiplying (1)(1) by 2x2x gives us: ∑n=1∞2nxn=2x(1−x)2→(3)\sum_{n=1}^{\infty} 2n x^n = \dfrac{2x}{(1-x)^2} \to (3)

Multiplying (2)(2) by x2x^2 gives us: ∑n=1∞n(n−1)xn=2x2(1−x)3→(4)\sum_{n=1}^{\infty} n(n-1)x^n = \dfrac{2x^2}{(1-x)^3} \to (4)

Now after adding (3)(3) and (4)(4) we get: ∑n=1∞(n2+n)xn=2x2(1−x)3+2x(1−x)2∑n=1∞(n2+n)xn=2x(1−x)3→(5)\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:

∑n=1P(n2+n)xn=∑n=1∞(n2+n)xn−∑n=P+1∞(n2+n)xn=2x(1−x)3−∑n=1∞((n+P)2+(n+P))xn+P=2x(1−x)3−xP[∑n=1∞(n2+n)xn+P∑n=1∞2nxn+(P2+P)∑n=1∞xn]=2x(1−x)3−xP[2x(1−x)3+2Px(1−x)2+(P2+P)x1−x]\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=12x = \dfrac{1}{2} gives us:

∑n=1Pn2+n2n=8−8+5P+P22P\sum_{n=1}^{P} \dfrac{n^2+n}{2^n} = 8 - \dfrac{8 + 5P + P^2}{2^P}

Ariel Gershon - 4 years, 8 months ago

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

Asif Hasan - 4 years, 7 months ago

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

∑n=1∞n2n+3n\large{\displaystyle \sum_{n=1}^{\infty} \frac{n}{2^n + 3^n}}

Asif Hasan - 4 years, 7 months ago

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

Ariel Gershon - 4 years, 7 months ago

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@Ariel Gershon – ohh :( :( thanks for the reply anyway!!

Asif Hasan - 4 years, 7 months ago

∑n=1Pn22n+∑n=1Pn2n\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 - 5 years, 11 months ago

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

Asif Hasan - 5 years, 11 months ago
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