Pointless Simplification
2 0 c ⋅ ⎝ ⎛ n = 1 ∏ 6 + e π i 5 ∫ 0 7 6 + ln ( e ) 1 d x ⎠ ⎞ ⋅ 4 n = 1 ∑ lo g c ( c 5 ) ( 4 1 ∫ − 2 2 d x ⋅ ( e π i + 2 ) ⋅ 5 ln ( e 1 6 c ) ⋅ ( i + 1 ) 8 1 )
Find the value of the expression above, if c is a non-zero real number.
The answer is 5.
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1 solution
Just in case somebody needs the LaTeX code:
\frac
{\displaystyle \sum_{n=1}^{\log_c(c^5)}
\left( \frac
{\ln (e^{16c}) \cdot \frac 1{(i+1)^8}}
{\frac 14 \int_{-2}^2 dx \cdot (e^{\pi i}+2) \cdot 5}
\right)}
{\frac c{20} \cdot
\left( \displaystyle \prod_{n=1}^{6+e^{\pi i}}
\sqrt[5]{ \int_0^7 \frac {1} {6+\ln(e)} dx }
\right) \cdot 4}
Faster method: 1. Note the factor of c/5 in the demonimator, giving a loose factor of 5 at the front of the expression 2. Look at the title 3. Assume everything else cancels out 4. Hooray! The answer is 5!
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Luckly that's not so easy in the new version, Nice method though.
How comes (i+1)^8=(2i)^4?
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( i + 1 ) 8 = ( ( i + 1 ) 2 ) 4 = ( i 2 + 2 ⋅ i ⋅ 1 + 1 2 ) 4 = ( − 1 + 2 i + 1 ) 4 = ( 2 i ) 4
x = 2 0 c ⋅ ⎝ ⎛ n = 1 ∏ 6 + e π i 5 ∫ 0 7 6 + ln ( e ) 1 d x ⎠ ⎞ ⋅ 4 n = 1 ∑ lo g c ( c 5 ) ( 4 1 ∫ − 2 2 d x ⋅ ( e π i + 2 ) ⋅ 5 ln ( e 1 6 c ) ⋅ ( i + 1 ) 8 1 ) = 5 c ⋅ ⎝ ⎛ n = 1 ∏ 5 5 ∫ 0 7 7 1 d x ⎠ ⎞ n = 1 ∑ 5 ( 4 5 ∫ − 2 2 1 d x 1 6 c ⋅ ( 2 i ) 4 1 ) = 5 c ⋅ ( n = 1 ∏ 5 5 1 ) n = 1 ∑ 5 ( 4 5 ⋅ 4 1 6 c ⋅ 1 6 1 ) = 5 c n = 1 ∑ 5 ( 5 c ) = 5 c 5 ⋅ ( 5 c ) = 5