Help Please (1 Viewer)

daweisun123 · May 2, 2017

Can you please help me with this questions

Drongoski · May 2, 2017

For (i) you can use the Binomial Series expansion of (1-x)^-2.

$(1-x)^{-2} = \sum _{i=0} ^{\infty} \binom {-2} i (-x)^{i} \\ \\ =\binom {-2} 0 (-x)^0 + \binom {-2} 1 (-x)^1 + \binom {-2} 2 (-x)^2 + \binom {-2} 3 (-x)^3 + \binom {-2} 4 (-x)^4 + \cdots \\ \\ = 1 + \frac {-2}{1} (-x) + \frac {{-2}\cdot {-3}}{1 \cdot 2} x^2 + \frac {{-2} \cdot {-3} \cdot {-4}}{1 \cdot 2 \cdot 3} (-x^3) + \frac {{-2}\cdot{-3} \cdot{-4}\cdot{-5}}{1 \cdot 2 \cdot3 \cdot 4} x^4 + \cdots \\ \\ = 1 + 2x + 3x^2 + 4 x^3 + 5 x^4 + \cdots$

Where is LaTeX??

Note that k+1Ck = k+1

Drongoski · May 2, 2017

If you are not familiar with the Binomial Series, you can use the usual Maclaurin expansion.

InteGrand · May 2, 2017

daweisun123 said:
View attachment 33912 Can you please help me with this questions

You can also do (i) by differentiating the geometric series 1/(1-x) = 1 + x + x^2 + x^3 + ..., for |x| < 1, differentiating the RHS term by term (recalling that a power series can be differentiated term-by-term within its interval of convergence).

To do the inductive step in (ii), differentiate both sides of the inductive hypothesis wrt x, again using term-by-term differentiation for the infinite series.

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Help Please (1 Viewer)

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