the coifficent of {r-1}th ,r th ,{r-1}th term in expansion of {x+1}n are in the ratio 1:3:5,find n&r.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The problem provides the equations that relates the binomial coefficients of the `(r-1),r ` and `(r+1)` terms, such that:

`(C_n^r)/(C_n^(r+1)) = 1/3 => (C_n^(r+1)) = 3(C_n^r)`

Using the factorial formulas yields:

`(n!)/((r+1)!(n - r - 1)!) = (3n!)/((r!)(n - r)!)`

Since `(r+1)! = r!(r+1)` and `(n-r)! = (n-r-1)!(n-r)` yields:

`1/(r+1) = 3/(n-r) => 3(r+1) = n-r`

`(C_n^(r+1))/(C_n^(r+2)) = 3/5`

`3(C_n^(r+2)) = 5(C_n^(r+1))`

`3(n!)/((r+2)!(n - r - 2!) = 5(n!)/((r+1)!(n - r - 1)!)`

`3/(r+2) = 5/(n-r-1 )=> 5(r+2) = 3(n-r-1)`

You need to solve for n and r the system of simultaneous equations, such that:

`{(3(r+1) = n-r),(5(r+2) = 3(n-r-1)):}`

`=> {(4r - n = -3),(8r - 3n = -13):} => {(-8r + 2n = 6),(8r - 3n = -13):} => -n = -7 => n = 7 r = (n-3)/4 => r = (7-3)/4 => r = 1`

Hence, evaluating r and n, under the given conditions, yields `r = 1` and `n = 7` .

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