Question

2
${2}^{2} + {4}^{2} + {6}^{2} + ..... {18}^{2} \: value$
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Answer 1

Step-by-step explanation:

We have,

${2}^{2} + {4}^{2} + {6}^{2} + ..... + {18}^{2}$

$= \sum^{9}_{n = 1}(2n)^{2}$

$= \sum^{9}_{n = 1} 4 {n}^{2}$

$= 4\sum^{9}_{n = 1} {n}^{2}$

$= 4. \frac{9(9 + 1)(2 \times 9 + 1)}{6}$

$= 4 \times \frac{9 \times 10 \times 19}{6}$

$= 1140$