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

Options

A. 964
B. 227
C. 232
D. 625

Answer

C

Explanation

Use general term of Binominal expansion `(x + a)^(n)` i.e., `T_(r + 1) = .^(n)C_(r ) ._(1)X^(n - 1) a^(r )`
Given binominal is `(x + 1)^(n)`, whose general term, is `T_(r + 1) = .^(n)C_(r ) X^(r )`.
According to the question, we have
`.^(n)C_(r ) : .^(n)C_(r ) : .^(n)C_(r + 1) = 2 : 15 : 70`
Now, `(.^(n)C_(r + 1))/(.^(n)C_(r )) = (2)/(15)`
`implies ((n !)/((r - 1)!(n - r + 1)!)/((n!)/(r!(n - r)!) = (2)/(15)`
`implies (r )/(n - r + 1) = (2)/(15) implies 15r = 2n - 2r + 2`
Similarly `(.^(n)C_(r ))/(.^(n)C_(r + 1)) implies ((n!)/(r!(n - r)!)/((n!)/((r + 1)!(n - r - 1)!) = (3)/(14)`
`implies (r + 1)/(n - r) = (3)/(14) implies 14 r + 14 = 3n - 3r`
On solving Eqs. (i) and (ii) we get
`n - 16 = 0 implies n = 16 and r = 2`
Now, the average = `(.^(10)C_(1) + .^(16)C_(2) + .^(15)C_(3))/(3)`
` = (16 + 120 + 560)/(3) = (696)/(3) = 232`