0

well ,i get this solution from wolframalpha and don't know where to start to prove it , or how to get this result.

$$\sum_{i=1}^N a^{N-i} *i = \frac{a^{N+1}-a*(N+1)+N}{(a-1)^2}$$

please help

Alberto
  • 11
  • Try k=N-i+1. It should be easier to sum. – herb steinberg Nov 21 '21 at 04:37
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 21 '21 at 04:37
  • The a^N is a constant term and can be factored to the left. i * a^-i looks a lot like the first derivative of a^-i with respect to a, a fact that could also be useful. – Paul Nov 21 '21 at 04:41

1 Answers1

0

i finally did it ;Paul, herb steinberg and Hans Lundmark thanks a lot guys. Here is the proof

$$\sum_{i=1}^N a^{N-i}*i= \sum_{i=0}^{N-1} a^{N-(i+1)}*(i+1) $$

$$=\sum_{i=0}^{N-1} a^{N-i-1}*(i+1)$$

$$= \sum_{i=0}^{N-1} a^{N-i-1}*i+ \sum_{i=0}^{N-1} a^{N-i-1} $$

$$= \frac{1}{a} * \sum_{i=0}^{N-1} a^{N-i}*i + \sum_{i=0}^{N-1} a^{N-i-1}$$

$$ = \frac{1}{a} * (\sum_{i=1}^{N-1} a^{N-i}*i + a^{N-0} * 0) + \sum_{i=0}^{N-1} a^{N-i-1} $$

$$= \frac{1}{a} * (\sum_{i=1}^{N-1} a^{N-i}*i) + \sum_{i=0}^{N-1} a^{N-i-1} $$

$$ = \frac{1}{a} * (a^{N-1}*1 +...+ a^{N-(N-1)}*(N-1)+0) + \sum_{i=0}^{N-1} a^{N-i-1}$$

$$= \frac{1}{a} * (a^{N-1}*1 +...+ a^{N-(N-1)}* (N-1)+ (a^{N-N}*N-a^{N-N}*N)) + \sum_{i=0}^{N-1} a^{N-i-1}$$

$$= \frac{1}{a} * ((a^{N-1}*1 +...+ a^{N-(N-1)}* (N-1)+ a^{N-N}*N)-a^{N-N}*N) + \sum_{i=0}^{N-1} a^{N-i-1}$$ $$= \frac{1}{a} * ((\sum_{i=1}^N a^{N-i}*i)-N) + \sum_{i=0}^{N-1} a^{N-i-1} $$

then we have

$$\sum_{i=1}^N a^{N-i}*i=\frac{1}{a} * (\sum_{i=1}^N a^{N-i}*i-N) + \sum_{i=0}^{N-1} a^{N-i-1} / * a$$

$$a*\sum_{i=1}^N a^{N-i}*i=\sum_{i=1}^N a^{N-i}*i-N + a*\sum_{i=1}^{N-1} a^{N-i-1} $$

$$a*\sum_{i=1}^N a^{N-i}*i=\sum_{i=1}^N a^{N-i}*i-N + a*\sum_{i=0}^{N-1} a^{N-i-1} / - \sum_{i=1}^N a^{N-i}*i $$

$$a*\sum_{i=1}^N a^{N-i}*i-\sum_{i=1}^N a^{N-i}*i= a*\sum_{i=0}^{N-1} a^{N-i-1} - N $$

$$a*\sum_{i=1}^N a^{N-i}*i-\sum_{i=1}^N a^{N-i}*i= a^{N-1+1}*\sum_{i=0}^{N-1} a^{-i} - N $$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= a^{N}*\frac{(\frac{1}{a})^{N}-1}{(\frac{1}{a})-1}-N$$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= a^{N}*\frac{((\frac{1}{a})^{N}-1)}{((\frac{1}{a})-1)}-N$$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= \frac{((\frac{a^N}{a^N})-a^{N})}{\frac{1-a}{a}}-N$$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= (a*\frac{(1-a^{N})}{(1-a)})-N$$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= (a*\frac{(a^{N}-1)}{(a-1)})-N$$

$$(a-1)*\sum_{i=1}^N a^{N-i}*i= \frac{a^{N+1}-a-N*a+N}{(a-1)}$$

$$\sum_{i=1}^N a^{N-i}*i= \frac{a^{N+1}-a-N*a+N}{(a-1)^2}$$

$$\sum_{i=1}^N a^{N-i}*i= \frac{a^{N+1}-a*(N+1)+N}{(a-1)^2}$$

Again thanks a lot.

Alberto
  • 11