If in the expansion of (1+x)n, the coefficients of three consecutive terms are 56,70,56, then the value of n and the position of the terms of these coefficients are given by
A
n=8, the terms are 4th,5th,6th
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B
n=7, the terms are 3rd,4th,5th
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C
n=8, the terms are 5th,6th,7th
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D
n=7, the terms are 4th,5th,6th
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Solution
The correct option is An=8, the terms are 4th,5th,6th Let the terms be nCr−1,nCr,nCr+1 Therefore from the given condition, we get nCr−1=56 nCr=70 nCr+1=56 Therefore nCr−1=nCr+1 Hence r−1=n−(r+1) r−1+r+1=n n=2r Now nCrnCr−1=54 Simplifying we get (2r)!r!r!×(r+1)!(r−1)!2r!=54 ...(n=2r) r+1r=54 Therefore r=4 Hence n=8. The terms are 8C3,8C4,8C5 =T4,T5,T6