Three types of functions in one!

Calculus Level 4

Find the value of i = 1 i 2 2 i i ! . \left\lfloor\displaystyle\sum_{i=1}^\infty\dfrac{i^22^i}{i!}\right\rfloor.


The answer is 44.

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2 solutions

Kartik Sharma
Feb 4, 2015

We know that e x = n = 0 x n n ! \displaystyle {e}^{x} = \sum_{n=0}^{\infty}{\frac{{x}^{n}}{n!}}

Differentiating w.rt. x x and then multiplying by x x ,

x e x = n = 0 n x n n ! \displaystyle x{e}^{x} = \sum_{n=0}^{\infty}{\frac{n{x}^{n}}{n!}}

Again differentiating w.r.t x x and multiplying by x x ,

x ( x e x + e x ) = n = 0 n 2 x n n ! \displaystyle x(x{e}^{x} + {e}^{x}) = \sum_{n=0}^{\infty}{\frac{{n}^{2}{x}^{n}}{n!}}

Substituting x = 2 x=2 ,

n = 0 n 2 2 n n ! = 6 e 2 \displaystyle \sum_{n=0}^{\infty}{\frac{{n}^{2}{2}^{n}}{n!}} = 6{e}^{2}

Lu Chee Ket
Feb 4, 2015

Floor (44.3343365935839) = 44

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