If 1 - 4x-1 - 1- 4x-1 - 2 n - 1- 4x-1 - 2 n - - a 0 - a 1 x- a 5 x 5- then n equals

Given #160; 1 /√( 4x+1 ) #160;{( 1+√( 4x+1 ) #160;/ 2 #160;) #160;^ n ( 1 √( 4x+1 ) #160;/ 2 #160;) #160;^ n } = a _ 0 + a _ 1 x+ ...

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Integration of Irrational Algebraic Fractions - 1

If 1 /√ 4x+...

Question

If $\frac{1}{\sqrt{4 x + 1}} [{(\frac{1 + \sqrt{4 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{4 x + 1}}{2})}^{n}] = a_{0} + a_{1} x + . . . + a_{5} x^{5}$ , then $n$ equals

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\frac{1}{\sqrt{4 x + 1}} [{(\frac{1 + \sqrt{4 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{4 x + 1}}{2})}^{n}] = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{5} x^{5}

n

\frac{1}{\sqrt{4 x + 1}} {{(\frac{1 + \sqrt{4 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{4 x + 1}}{2})}^{n}} = a_{0} + a_{1} x + . . . . a_{5} x^{5}

n =

\frac{1}{\sqrt{4 x + 1}} [{(\frac{1 + \sqrt{4 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{4 x + 1}}{2})}^{n}]

a_{0} + a_{1} x + . . . + a_{5} x^{5}

(1 + x) (1 + x + x^{2}) . . . . . (1 + x + . . . . . + x^{n}) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . .

a_{0} + a_{2} + a_{4} + . . . . .

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{40}

a_{1} + a_{3} + a_{5} + . . . . . . + a_{39}

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