Maybe self-destructive

Calculus Level 5

( ∑ x = 0 ∞ ( 2 x ) ! ( 5 x + 6 x ) x ! 2 ⋅ 3 0 x ) 2 \large \left (\sum_{x=0}^\infty \dfrac{ (2x)! (5^x + 6^x) }{x!^2 \cdot 30^x } \right )^2

If the value of the expression above is in the form of a + b a + \sqrt b , where a a and b b are positive integers, find a + b a+b .


The answer is 68.

Joel Yip by Joel Yip , 21, Singapore

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1 solution

Joel Yip
Feb 14, 2016

∑ n = 0 ∞ ( 2 n ) ! ( 5 x + 6 x ) n ! 2 30 n = ∑ n = 0 ∞ ( 2 n n ) ( 1 5 n + 1 6 n ) \sum _{ n=0 }^{ \infty }{ \frac { \left( 2n \right) !\left( { 5 }^{ x }+6^{ x } \right) }{ { n! }^{ 2 }{ 30 }^{ n } } =\sum _{ n=0 }^{ \infty }{ \left( \begin{matrix} 2n \\ n \end{matrix} \right) \left( \frac { 1 }{ { 5 }^{ n } } +\frac { 1 }{ { 6 }^{ n } } \right) } } ,

so, ∑ n = 0 ∞ ( 2 n n ) ( 1 5 n + 1 6 n ) = ∑ n = 0 ∞ ( 2 n n ) 5 n + ∑ n = 0 ∞ ( 2 n n ) 6 n = 1 1 − 4 5 + 1 1 − 4 6 = 1 1 5 + 1 1 3 = 5 + 3 { \sum _{ n=0 }^{ \infty }{ \left( \begin{matrix} 2n \\ n \end{matrix} \right) \left( \frac { 1 }{ { 5 }^{ n } } +\frac { 1 }{ { 6 }^{ n } } \right) } =\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ { 5 }^{ n } } } }+\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ 6^{ n } } } \\ =\frac { 1 }{ \sqrt { 1-\frac { 4 }{ 5 } } } +\frac { 1 }{ \sqrt { 1-\frac { 4 }{ 6 } } } \\ =\frac { 1 }{ \sqrt { \frac { 1 }{ 5 } } } +\frac { 1 }{ \sqrt { \frac { 1 }{ 3 } } } \\ =\sqrt { 5 } +\sqrt { 3 }

( 5 + 3 ) 2 = 8 + 2 15 = 8 + 60 { \left( \sqrt { 5 } +\sqrt { 3 } \right) }^{ 2 }\\ =8+2\sqrt { 15 } \\ =8+\sqrt { 60 }

thus the answer is 68

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