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Question

# If some three consecutive coefficients in the binomial expansion of (x+1)n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is :

A
227
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B
232
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C
625
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D
964
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Solution

## The correct option is B 232Let the coefficients of terms tr, tr+1, tr+2 are in ratio 2:15:70. ∴ nCr−1: nCr: nCr+1=2:15:70 ⇒ nCr−1 nCr=215⇒rn−r+1=215 ⇒15r=2n−2r+2 ⇒17r=2n+2 …[1] Also, nCr nCr+1=1570⇒r+1n−r=1570 ⇒14r+14=3n−3r ⇒17r=3n−14 …[2] From [1] and [2], we get n=16 and r=2 First term = 16C1=16 Therefore, other two terms are 16C2=120 and 16C3=560. Therefore, average of coefficients =232

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