Evenly Odd , Oddly Even

Algebra Level 3

Find the sum of the series up to n n terms

1 2 + ( 2 × 2 2 ) + 3 2 + ( 2 × 4 2 ) + ⋯ + n 2 1^2 + (2 \times 2^2) + 3^2+ (2 \times 4^2)+\cdots +n^2 .

Where n n is an odd number.

i.e Evaluate

∑ m = 1 n ( ( 2 m − 1 ) 2 + [ 2 × ( 2 m ) 2 ] ) \large \displaystyle {\sum_{m = 1}^ {n} \left ( (2m-1)^2+[2 \times(2m)^2] \right ) }

n ( n − 1 ) 2 \dfrac {n(n-1)}{2} ( n + 1 ) n 2 2 \dfrac {(n+1)n^2}{2} n ( n + 1 ) 2 \dfrac {n(n+1)}{2} ( n − 1 ) n 2 2 \dfrac {(n-1)n^2}{2}

Vaibhav Gupta by Vaibhav Gupta , 22, India

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1 solution

First, you haven't written the sum correctly. Actually, the sum we have to evaluate is

∑ m = 1 n − 1 2 [ ( 2 m + 1 ) 2 + 2 ⋅ ( 2 m ) 2 ] + 1 \sum _{ m=1 }^{ \frac { n-1 }{ 2 } }{ \left[ { \left( 2m+1 \right) }^{ 2 }+{ 2\cdot \left( 2m \right) }^{ 2 } \right] } +1

Operating and expanding, this is equal to

[ 12 ⋅ ∑ m = 1 n − 1 2 m 2 ] + [ 4 ⋅ ∑ m = 1 n − 1 2 m ] + n − 1 2 + 1 \left[ 12\cdot \sum _{ m=1 }^{ \frac { n-1 }{ 2 } }{ { m }^{ 2 } } \right] +\left[ 4\cdot \sum _{ m=1 }^{ \frac { n-1 }{ 2 } }{ m } \right] +\frac { n-1 }{ 2 } +1

As we know the formulas for the first sums

12 ( n − 1 2 ) ( n − 1 2 + 1 ) n 6 + 4 ( n − 1 2 ) ( n − 1 2 + 1 ) 2 + n − 1 2 + 1 \frac { 12\left( \frac { n-1 }{ 2 } \right) \left( \frac { n-1 }{ 2 } +1 \right) n }{ 6 } +\frac { 4\left( \frac { n-1 }{ 2 } \right) \left( \frac { n-1 }{ 2 } +1 \right) }{ 2 } +\frac { n-1 }{ 2 } +1

Finally, this is equal to n 2 ( n + 1 ) 2 \frac { { n }^{ 2 }(n+1) }{ 2 }

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