1.

The sum of the coefficients of the first three terms in the expansion of`(x-3/(x^2))^m ,x!=0,`m being a natural number, is 559. Find the term of the expansion containing `x^3`.

Answer» representing as sum of first three terms
`.^mC_0(-3)^0 + .^mC_1(-3)^1 + .^mC_2(-3)^2 = 559`
`1 + m(-3) + (m(m-1))/2*9 = 559`
`-3m + (9m^2-9m)/2 = 558`
`3m^2/2- 5m/2 = 186`
`3m^2 - 5m -372 = 0`
`m = (5+- sqrt(25+12*372))/6`
`= (5 + sqrt(4464+25))/6`
`= (5+67)/6 72/6 = 12`
to find the term for `x^3`
so,`m-3r=3`
`r=3`
so fourth term will be taken`t_4 = .^12C_3(-3)x^3`
`= 12*11*10/(3*2)* (-27x^3)`
`= -5940x^3`


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