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Question

If r is a fixed positive integer, then if we use induction we can find that the expression (r+1)(r+2)(r+3)....(r+n) is always divisible by

A
n+1
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B
nn
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C
n!
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D
n2+1
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Solution

The correct option is C n!

Let P(n): (r+1)(r+2)(r+3)....(r+n)=n!.k,

Where k is an integer.

When n=1,r+1=1!.(r+1)

P(1) is true.

let P(m) be true,i.e.,

(r+1)(r+2)(r+3)....(r+m)=m!.k ....(1)

Now, (r+1)(r+2)(r+3)....(r+m)(r+m+1)

= r(r+1)(r+2)(r+3)....(r+m)+(m+1)(r+1)(r+2)(r+3)...(r+m)

=(r+m)!(r1)!+(m+1)(m!)k,using(1)

=(m+1)!.(r+m)!(r1)!(m+1)!+(m+1)!.k

=(m+1)!.(r+mCr1+k)

=(m+1)!(integer+k)

P(m+1) is true P(m+1)is true.

P(n) is true for all nϵN.


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