Question

Find the last digit of the number 1^2+2^2+...+99^2

Answer 1

Using formula for sum of the squares of n natural numbers,
${1}^{2} + {2}^{2} + .... + {99}^{2} \\ = \frac{99 \times 100 \times 199}{6} = 328350$
Last digit is 0

Answer 2

The last digit of the number 1² + 2² + . . . + 99² is 0

Given :

The number 1² + 2² + . . . + 99²

To find :

The last digit of the number 1² + 2² + . . . + 99²

Formula :

$\displaystyle \sf{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6} }$

Solution :

Step 1 of 2 :

Calculate the sum

We know that

$\displaystyle \sf{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6} }$

Thus we get

$\displaystyle \sf{ {1}^{2} + {2}^{2} + ... + {99}^{2}}$

$\displaystyle \sf{ = \frac{99 \times (99 + 1) \times \{(2 \times 99) + 1 \}}{6} }$

$\displaystyle \sf{ = \frac{99 \times 100 \times 199}{6} }$

$\displaystyle \sf{ = \frac{1970100}{6} }$

$\displaystyle \sf{ = 328350 }$

Step 2 of 2 :

Find the last digit of the number

Since 1² + 2² + . . . + 99² = 328350

Hence last digit of the number 1² + 2² + . . . + 99² is 0

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