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Question Number 207262 by hardmath last updated on 10/May/24

$a_{n} - number series$ $a_{1} = 5$ $d = 3$ $a_{2}^{2} - a_{1}^{2} + a_{4}^{2} - a_{3}^{2} + a_{6}^{2} - a_{5}^{2} + . . . + a_{10}^{2} - a_{9}^{2} = ?$
	Answered by A5T last updated on 10/May/24

	$(a_{2} - a_{1}) (a_{2} + a_{1}) = d \times [2 a_{1} + d]$ $a_{2 n}^{2} - a_{2 n - 1}^{2} = (a_{2 n} - a_{2 n - 1}) (a_{2 n} + a_{2 n - 1})$ $= d \times [a_{1} + (2 n - 1) d + a_{1} + (2 n - 2) d]$ $= d \times [2 a_{1} + d (4 n - 3)]$ $\underset{n = 1}{\sum^{5}} a_{2 n}^{2} - a_{2 n - 1}^{2} = d [10 a_{1} + d (1 + 5 + 9 + 13 + 17)]$ $= 3 [50 + 3 (45)] = 555$
	Commented by hardmath last updated on 10/May/24

	$perfect dear professor thankyou$
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