Dear all, how can I prove the following summation (See attachment)?
I tried to use integrals but the result was kind of broad.
Attachment 29481
Dear all, how can I prove the following summation (See attachment)?
I tried to use integrals but the result was kind of broad.
Attachment 29481
$\displaystyle 0 = \sum_{k = 0}^0 ka^k = \dfrac{a}{(1-a)^2}[1-(0+1)a^0 + 0a^{0+1}] = 0$
Assume the formula is correct for all nonnegative integers up to $\displaystyle n$.
$\displaystyle \sum_{k=0}^{n+1}ka^k = \sum_{k=0}^n ka^k + (n+1)a^{n+1}$
By the induction assumption, we have
$\displaystyle \sum_{k=0}^n ka^k + (n+1)a^{n+1} = \dfrac{a}{(1-a)^2}[1-(n+1)a^n + na^{n+1}] + (n+1)a^{n+1}$
Can you finish from here?
The original formula? Or the formula I wrote: $\displaystyle \dfrac{1-a^{n+2}}{1-a} = \sum_{k = 0}^{n+1}a^k$? Because that formula is just the standard geometric sum:
$\displaystyle (1+a+a^2+\ldots+a^{n+1})(1-a)=$
$\displaystyle \begin{array}{cccccccccccc} & 1 & + & a & + & a^2 & + & \ldots & + & a^{n+1} & & \\ & & - & a & - & a^2 & - & \ldots & - & a^{n+1} & - & a^{n+2} \\ = & 1 & & & & & & & & & - & a^{n+2}\end{array}$