If (1+x)n=n∑r=0arxr. Then (1+a1a0)(1+a2a1)…(1+anan−1) is equal to
A
(n+1)n+1n!
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B
(n+1)nn!
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C
nn−1(n−1)!
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D
(n+1)n−1(n−1)!
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Solution
The correct option is B(n+1)nn! arar−1=nCrnCr−1=n−r+1r Now, 1+arar−1=1+n−r+1r⇒1+arar−1=n+1r So, (1+a1a0)(1+a2a1)…(1+anan−1)=r=n∏r=1(1+arar−1)=r=n∏r=1(n+1r)=(n+1)nr=n∏r=1(1r)=(n+1)nn!