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-

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} + (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 + \dots . 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} + \dots 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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