Adding Back In a Term into a Summation

[r_yackel]

\(\displaystyle (x+2)\sum_{n=2}^{\infty}C_n n(n-1)x^{n-2}\)

Need help on getting the (x+2) back into the summation.
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Edited by stapel -- Reason for edit: making requested correction

 

[stapel]

\(\displaystyle (x+2)\sum_{n=2}^{\infty}C_n n(n-1)x^{n-2}\)

Need help on getting the (x+2) back into the summation.

As the expression stands, the (x + 2) will be multiplied onto the summation. The Distributive Law (from back in pre-algebra) says that this multiplication "distributes over addition", so the (x + 2) will become a factor of every term in the summation.

So move the factor in:

. . . . .\(\displaystyle \sum_{n=2}^{\infty}\,C_n\, \left(n(n\,-\,1)\right)\,(x\, +\, 2)\,x^{n-2}\)



Eliz.