Question

1cube+2cube+3cube+.......+ncube whole cube​

Answer 1

Answer:

³ + 3³ + 5³ + ... + (2n - 1)³

By Adding and subtracting

2³ + 4³ + 6³ +... + (2n)³ We get

1³ + 2³ + 3³ + ... + (2n-1)³ + (2n)³

- (2³ + 4³ + 6³ + ... + (2n)³)

= [ 2n(2n+1)/2 ]² - 2³[ 1³ + 2³ + 3³+ ... + (2n)³]

= [ n(2n + 1) ]² - 2³ [ n(n + 1)/2 ]²

= n²(2n+1)² - 2³ n²(n + 1)²/2²

= n²(2n + 1)² - 2n²(n + 1)²

= n² [ (2n + 1)² - 2(n + 1)² ]

= n² [ 4n² + 4n + 1 - 2n² - 4n - 2 ]

= n² (2n² - 1)

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★ Formula Used :

Sum of cube of first n natural number

= [ n(n + 1)/2 ]²

Answer 2

Answer:

${{ \frac{ {n}^{2} (n + 1)}{4} }}^{2}$

Step-by-step explanation:

${1}^{3} + {2}^{3} + {3}^{3} + {4}^{3} + ...... {n}^{3} \\ formula \: \\ \ {{ \frac{ {n}^{2} (n + 1)}{4} }}^{2}$