Question

12. Find the sum of the cubes of first 12 natural numbers.​

Answer 1

Answer:

6084

Step-by-step explanation:

1+8+27+64+125+216+343+512+729

+1,000+1,331+1,728 = 6084

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Answer 2

The sum of the cubes of first 12 natural numbers = 6084

Given :

The first 12 natural numbers

To find :

The sum of the cubes of first 12 natural numbers

Formula :

$\displaystyle \sf{ {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3} = {\bigg[ \frac{n(n + 1)}{2} \bigg]}^{2} }$

Solution :

Step 1 of 2 :

Write down the given numbers

Here we are given first 12 natural numbers

Step 2 of 2 :

Find the sum of the cubes of first 12 natural numbers

We know that ,

$\displaystyle \sf{ {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3} = {\bigg[ \frac{n(n + 1)}{2} \bigg]}^{2} }$

Now we have to calculate sum of the cubes of first 12 natural numbers

∴ n = 12

$\displaystyle \sf{ {1}^{3} + {2}^{3} + {3}^{3} + ... + {12}^{3} }$

$\displaystyle \sf{ = {\bigg[ \frac{12(12 + 1)}{2} \bigg]}^{2} }$

$\displaystyle \sf{ = {\bigg[ \frac{12 \times 13}{2} \bigg]}^{2} }$

$\displaystyle \sf{ = {(6 \times 13)}^{2} }$

$\displaystyle \sf{ = {78}^{2} }$

$\displaystyle \sf{ = 6084}$

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