Consider the binomial expansion of (1+7X)^23 in accending powers of x. Given that x is positive, find the range of values of x for which the greatest term of the expansion is the fouth term.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the formula that gives the general term, such that:

`T_(k+1) = C_n^k a^(n - k)*b^k`

Identifying `a = 1, b = 7x, n = 23, k+1 = 4` yields:

`T_4 = C_23^3*1^(23 - 3)*(7x)^3`

`T_4 = (23!)/(3!(23-3)!)*1^(23 - 3)*(7x)^3`

`T_4 = (23!)/(3!20!)*1^(23 - 3)*(7x)^3`

`T_4 = (20!21*22*23)/(1*2*3*20!)*(7x)^3`

Reducing duplicate factors ,yields:

`T_4 = 7*11*23*(7x)^3`

The problem provides the information that T_4 is the greatest term of binomial expansion, hence, you may considering the following inequality, such that:

`T_4 > T_3`

Evaluating `T_3` yields:

`T_3 = C_23^2*1^(23 - 2)*(7x)^2`

`T_3 = (23!)/(2!21!)*1^(23 - 2)*(7x)^2`

`T_3 = (21!22*23)/(2*21!)*1^(23 - 2)*(7x)^2`

`T_3 = 11*23*(7x)^2`

Replacing the expressions of `T_3` and `T_4` in inequality, yields:

`7*11*23*(7x)^3 > 11*23*(7x)^2`

Reducing duplicate terms yields:

`7*(7x)^3 > (7x)^2 => 7*(7x)^3 - (7x)^2 > 0`

You need to attach the corresponding equation, such that:

`7*(7x)^3 - (7x)^2 = 0 => (7x)^2*(7*7x - 1) = 0`

Using the zero product rule yields:

`(7x)^2 = 0 => x = 0`

`49x - 1 = 0 => 49x = 1 => x = 1/49`

You should notice that the inequality `7*(7x)^3 - (7x)^2 > 0` holds for `x < 0` or `x > 1/49.`

Since the problem provides the information that `x > 0` , hence, you need to only keep the interval x in (1/49,oo).

Hence, evaluating the range of values of x, under the given conditions, yields `x in (1/49,oo).`

x in (1/49,oo).

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