Question

Evaluate the following sum without simplify the cubes : $1^{3}- 2^{3}+3^{3}-4^{3}+...+9^{3}$

Use sequence and series to solve this.

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: {1}^{3} - {2}^{3} + {3}^{3} - {4}^{3} + ... + {9}^{3}$

can be rewritten as

$\rm \: = \: {1}^{3} + {3}^{3} + {5}^{3} + {7}^{3}+ {9}^{3} - ( {2}^{3} + {4}^{3} + {6}^{3} + {8}^{3})$

can be further rewritten as

$\rm \: = \: ({1}^{3} + {2}^{3} + {3}^{3} +... + {9}^{3}) - 2( {2}^{3} + {4}^{3} + {6}^{3} + {8}^{3})$

can be further rewritten as

$\rm \: = \: ({1}^{3} + {2}^{3} + {3}^{3} +... + {9}^{3}) - 2( {(1.2)}^{3} + {(2.2)}^{3} + {(2.3)}^{3} + {2.4)}^{3})$

$\rm \: = \: ({1}^{3} + {2}^{3} + {3}^{3} +... + {9}^{3}) - 2. {2}^{3} ( {1}^{3} + {2}^{3} + {3}^{3} + {4}^{3})$

$\rm \: = \: \displaystyle\sum_{k=1}^9\rm {k}^{3} - 16\displaystyle\sum_{k=1}^4\rm {k}^{3}$

We know,

$\boxed{ \rm{ \:\displaystyle\sum_{k=1}^n\rm {k}^{3} \: = \: {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} \: }}$

So, using this result, we get

$\rm \: = \: \bigg(\dfrac{9(9 + 1)}{2}\bigg)^{2} - 16 \bigg(\dfrac{4(4 + 1)}{2}\bigg)^{2}$

$\rm \: = \: {45}^{2} - 16 \times {(10)}^{2}$

$\rm \: = \: 2025 - 1600$

$\rm \: = \: 425$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\bf \: {1}^{3} - {2}^{3} + {3}^{3} - {4}^{3} + ... + {9}^{3} = 425 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:\displaystyle\sum_{k=1}^n\rm k \: = \: \frac{n(n + 1)}{2} \: }}$

$\boxed{ \rm{ \:\displaystyle\sum_{k=1}^n\rm {k}^{2} \: = \: \frac{n(n + 1)(2n + 1)}{6} \: }}$