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Question: Let a<sub>k</sub> = (k<sup>2</sup> + 1) k !&b<sub>k</sub> = a<sub>1</sub> + a<sub>2</sub> + a<sub>3<...

Let ak = (k2 + 1) k !&bk = a1 + a2 + a3 …. + ak. If a100b100\frac{a_{100}}{b_{100}} = mn\frac{m}{n} then n – m equals

A

100

B

99

C

98

D

None

Answer

99

Explanation

Solution

ak = (k2 + 1) k!

= (k (k + 1) – (k –1)) k!

= k (k + 1) k! – (k –1) k!

= k (k + 1) ! – (k – 1) k!

\ a1 = 2! – 0

a2 = 2.3! – 2!

a3 = 3. 4! – 2. 3!

ak = k (k + 1) –(k –1)k!

Adding

a1 + a2 …… +ak = k (k + 1)!

̃ bk = k (k + 1)!

\ akbk\frac{a_{k}}{b_{k}} = (k2+1)k!k(k+1)!\frac{(k^{2} + 1)k!}{k(k + 1)!}

akbk\frac{a_{k}}{b_{k}} = k2+1k(k+1)\frac{k^{2} + 1}{k(k + 1)}

\ a100b100\frac{a_{100}}{b_{100}} = 1000110100\frac{10001}{10100} = mn\frac{m}{n}

\ n – m = 99