Given that the sequence {a_n } is defined as a_1 =2, and a_(n+1) =a_n +(2n−1) for all n≥1. Find the last two digits of a


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Question Number 115133 by ZiYangLee last updated on 23/Sep/20
$Given that the sequence {a_n } is defined as a_1 =2, and a_(n+1) =a_n +(2n−1) for all n≥1. Find the last two digits of a_(100) .$
$Given that the sequence {a_{n}} is defined$ $as a_{1} = 2, and a_{n + 1} = a_{n} + (2 n - 1) for all n ⩾ 1 .$ $Find the last two digits of a_{100} .$
	Answered by Olaf last updated on 23/Sep/20

	$Let S_{n} = \underset{k = 1}{\sum^{n}} a_{k}$ $S_{n + 1} - S_{n} = \underset{k = 1}{\sum^{n + 1}} a_{k} - \underset{k = 1}{\sum^{n}} a_{k}$ $S_{n + 1} - S_{n} = a_{1} + \underset{k = 1}{\sum^{n}} a_{k + 1} - \underset{k = 1}{\sum^{n}} a_{k}$ $S_{n + 1} - S_{n} = a_{1} + \underset{k = 1}{\sum^{n}} (a_{k + 1} - a_{k})$ $S_{n + 1} - S_{n} = 2 + \underset{k = 1}{\sum^{n}} (2 k - 1)$ $S_{n + 1} - S_{n} = 2 + (2 \underset{k = 1}{\sum^{n}} k) - n$ $S_{n + 1} - S_{n} = 2 + 2 \frac{n (n + 1)}{2} - n$ $S_{n + 1} - S_{n} = n^{2} + 2$ $But S_{n + 1} - S_{n} = a_{n + 1}$ $\Rightarrow a_{n + 1} = n^{2} + 2$ $a_{100} = 99^{2} + 2$ $a_{100} = 10000 - 200 + 1 + 2 = 9803$ $Two last digits are 03$
	Answered by Dwaipayan Shikari last updated on 23/Sep/20

	$a_{n + 1} = a_{n} + (2 n - 1)$ $a_{2} = a_{1} + (2 - 1) \Rightarrow a_{2} = 3$ $a_{3} = 3 + (4 - 1) = 6$ $a_{100} = a_{99} + (2.99 - 1)$ $a_{100} = (a_{98} + 2.98 - 1) + 2.99 - 1$ $a_{100} = a_{1} + 2 (1 + 2 + . . . . 99) - 99$ $a_{100} = 2 + 99.100 - 99$ $a_{100} = 2 + 99^{2} = 9803$
	Answered by Bird last updated on 24/Sep/20

	$a_{n + 1} - a_{n} = 2 n - 1 \Rightarrow$ $\sum_{k = 1}^{n - 1} (a_{k + 1} - a_{k}) = \sum_{k = 1}^{n - 1} (2 k - 1) \Rightarrow$ $a_{2} - a_{1} + a_{3} - a_{2} + . . . a_{n} - a_{n - 1}$ $= 2 \sum_{k = 1}^{n - 1} k - (n - 1)$ $= 2 . \frac{(n - 1) n}{2} - n + 1 = n^{2} - n - n + 1$ $= n^{2} - 2 n + 1 \Rightarrow a_{n} = n^{2} - 2 n + 1 + 2$ $= n^{2} - 2 n + 3 \Rightarrow a_{100} = 100^{2} - 2.100 + 3$ $= 10000 - 200 + 3$ $= 9803$
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