Sum

Calculus Level 2

∑ n = 0 ∞ 2 n + 3 n n ! = ? \large \sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ n }+{ 3 }^{ n } }{ n! } } = \, ?

1 e 2 + e 3 { e }^{ 2 }+{ e }^{ 3 } e 2 + e { e }^{ 2 }+{ e } e 5 { e }^{ 5}

Ali Math by Ali Math , 45, Saudi Arabia

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2 solutions

We have e x = ∑ n = 0 ∞ x n n ! \large e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} , ∑ n = 0 ∞ 2 n + 3 n n ! = ∑ n = 0 ∞ 2 n n ! + ∑ n = 0 ∞ 3 n n ! = e 2 + e 3 \large \sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ n }+{ 3 }^{ n } }{ n! } } = \sum_{n=0}^{\infty} \frac{2^n}{n!} + \sum_{n=0}^{\infty} \frac{3^n}{n!} = e^2+e^3

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Nice solution

Jun Arro Estrella - 5 years, 1 month ago

For series representation e 2 ( 1 + e ) = ( ∑ k = 0 ∞ − 1 + k + z k ! ) 2 ( z + ∑ k = 0 ∞ − 1 + k + z k ! ) z 3 e^2(1+e)=\dfrac{\left(\sum_{k=0}^{\infty} \dfrac{-1+k+z}{k!}\right)^2\left(z+\sum_{k=0}^{\infty}\dfrac{-1+k+z}{k!}\right)}{z^3}

Partial sum formula:

∑ n m 2 n + 3 n n ! = e 2 ( Γ ( m + 1 , 2 ) + e Γ ( m + 1 , 3 ) ) Γ ( m + 1 ) \sum_{n}^{m}\frac{2^n+3^n}{n!} = \frac{e^2(\Gamma(m+1,2)+e\Gamma(m+1,3))}{\Gamma(m+1)}

ADIOS!!! \LARGE \text{ADIOS!!!}

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