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Question

Let Tn=(n2+1)n! and Sn=T1+T2+T3++Tn. If T10S10=ab, where a and b are relatively prime natural numbers, then the value of (ba) is

A
9
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B
2
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C
13
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D
14
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Solution

The correct option is A 9
Tn=(n2+1)n!
=[(n+1)n(n1)]n!
Tn=n(n+1)!(n1)n!

T1=1×2!0T2=2×3!1×2!T3=3×4!2×3!T4=4×5!3×4! Tn=n(n+1)!(n1)n!

Sn=T1+T2+T3++Tn
Sn=n(n+1)!
Now, T10S10=101×10!10×11!=101110
ba=110101=9

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