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STEP III 2007 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 2 solution

(i)

\displaystyle 1.3.5.7...(2n-1) = \frac{1.2.3.4...(2n-1)(2n)}{2.4.6.8...(2n)}

\displaystyle = \frac{(2n)!}{(2.1)(2.2)(2.3)(2.4)...(2.n)} = \frac{(2n!)}{2^nn!} as required.

\displaystyle \frac{1}{\sqrt{1-4x}} = (1-4x)^{-\frac{1}{2}}
Binomial expansion valid for \displaystyle |4x| < 1 i.e. \displaystyle |x| < \frac{1}{4} as given.

\displaystyle \frac{1}{\sqrt{1-4x}} = 1 + 2x + ... + \frac{(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2r-1}{2}).(-4x)^r}{r!} + ...

Every successive term is multiplied by two extra negative terms, one fraction from the exponent and one power of -4x, so every term stays positive overall, so all minus signs can be removed.

\displaystyle = 1 + 2x + ... + \frac{(\frac{1}{2})(\frac{3}{2})...(\frac{2r-1}{2}).(4x)^r}{r!} + ...

Using the formula just proved and noting that there is a divisor of \displaystyle 2^r in total from the product of \displaystyle (\frac{3}{2})...(\frac{2r-1}{2}):

\displaystyle = 1 + 2x + ... + \frac{(2r!)4^rx^r}{2^r2^r(r!)^2} + ...

\displaystyle = 1 + 2x + ... + \frac{(2r!)x^r}{(r!)^2} + ...

\displaystyle = 1 + \sum_1^{\infty}\frac{(2n!)x^n}{(n!)^2} as required.

Verify that 2x is the first term by setting n = 1 and getting 2x.

(ii)

Differentiate both sides with respect to x:

\displaystyle \frac{2}{(1-4x)^{\frac{3}{2}}} = \sum_1^{\infty}\frac{n(2n!)x^{n-1}}{(n!)^2}

Setting \displaystyle x = \frac{6}{25}:

\displaystyle 250 = \sum_1^{\infty}\frac{n(2n!)\frac{6}{25}^{n-1}}{(n!)^2}
Multiply both sides by \displaystyle \frac{6}{25}:

\displaystyle \sum_1^{\infty}\frac{n(2n!)\frac{6}{25}^n}{(n!)^2} = 60 as required.

(iii)

This time integrate instead:

\displaystyle -\frac{1}{2}\sqrt{1-4x} = x + \sum_1^{\infty}\frac{(2n!)x^{n+1}}{(n+1)(n!)^2} + C
Set x = 0:

\displaystyle -\frac{1}{2} = 0 + 0 + C \rightarrow C = -\frac{1}{2}

Set x = \displaystyle \frac{2}{9}:

\displaystyle -\frac{1}{6} = \frac{2}{9} + \sum_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} - \frac{1}{2}

\displaystyle \sum_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n+2}(n+1)!(n!)} = \frac{1}{9}

\displaystyle \sum_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} = 1 as required.

Solution by Speleo.

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