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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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