Question

Find the value of 1 square + 2 square + 3 square + .... +10 square and hence deduce 2 square + 4 square + 6square+ .... + 20square

Answer 1

Answer:

so basically,

Step-by-step explanation:

(1²+2²+3²+4²+5²+6²+7²+8²+9²+10²) - (2²+4⁴+6⁶+8²+10²+12²+14²+16²+18²+20²)

= (55²) - (110²)

= 55²-110²

= -55²

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Answer 2

• 1² + 2² + 3² + . . . + 10² = 385

• 2² + 4² + 6² + . . . + 20² = 1540

Given :

1² + 2² + 3² + . . . + 10²

To find :

• The value of 1² + 2² + 3² + . . . + 10²

• Hence deduce 2² + 4² + 6² + . . . + 20²

Formula Used :

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

Solution :

Step 1 of 2 :

Find value of 1² + 2² + 3² + . . . + 10²

We know that ,

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

Thus we get

$\displaystyle \sf{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {10}^{2} }$

$\displaystyle \sf{= \frac{10 \times ( 10 + 1)\times (20 + 1)}{6} }$

$\displaystyle \sf{= \frac{10 \times 11 \times 21}{6} }$

$\displaystyle \sf{= 385}$

Step 2 of 2 :

Deduce the value of 2² + 4² + 6² + . . . + 20²

$\displaystyle \sf {2}^{2} + {4}^{2} + {6}^{2} + ... + {20}^{2}$

$\displaystyle \sf = {2}^{2} \times ({1}^{2} + {2}^{2} + {3}^{2} + ... + {10}^{2} )$

$\displaystyle \sf = 4 \times 385\: \: \: \bigg[ \: \because \: {1}^{2} + {2}^{2} + {3}^{2} + ... + {10}^{2} = 385\bigg]$

$\displaystyle \sf = 1540$

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