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I have the following series and I am unable to figure out which series it belongs to and how to solve it

$a(x+1)+a^{2}(x+2)+…+a^{N}(x+N)$

Above series is a generalization of my actual series

$\dfrac{1}{2}(x+1)+\dfrac{1}{4}(x+2) +\cdots +\dfrac{1}{2^{N}}(x+N)$

robjohn
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user4131265
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3 Answers3

1

Hint:

Note that $$ \begin{align} (1-a)\sum_{k=1}^na^k &=\sum_{k=1}^na^k-\sum_{k=1}^na^{k+1}\\ &=\sum_{k=1}^na^k-\sum_{k=2}^{n+1}a^k\\ &=a-a^{n+1}\tag{1} \end{align} $$ Therefore, $$ \sum_{k=1}^na^k=\frac{a-a^{n+1}}{1-a}\tag{2} $$ Similarly note that $$ \begin{align} (1-a)\sum_{k=1}^nka^k &=\sum_{k=1}^nka^k-\sum_{k=1}^nka^{k+1}\\ &=\sum_{k=1}^nka^k-\sum_{k=2}^{n+1}(k-1)a^k\\ &=\sum_{k=1}^na^k-na^{n+1}\\ &=\frac{a-a^{n+1}}{1-a}-na^{n+1}\\ &=\frac{a-(n+1)a^{n+1}+na^{n+2}}{1-a}\tag{3} \end{align} $$ Therefore, $$ \sum_{k=1}^nka^k=\frac{a-(n+1)a^{n+1}+na^{n+2}}{(1-a)^2}\tag{4} $$ Formulas $(2)$ and $(4)$ should help.

robjohn
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0

We write the series as

$$ \sum_{k=1}^{N} (x+k)a^k = x\sum_{k=1}^{N} {a^k}+ \sum_{k=1}^{N} {k}{a^k} \longrightarrow (1). $$

You can use the identity

$$ \sum_{k=1}^{N} t^k = \frac{t^{N+1}-t}{t-1} \longrightarrow (*)$$

to sum the series in $(1)$.

Added: The first series in $(1)$ is straightforward just substitute $t=a$ in $(*)$ as

$$ x\sum_{k=1}^{N} {a^k} = x\frac{a^{N+1}-a}{a-1} .$$

For the second series you just need to differentiate $(*)$ as

$$ \sum_{k=1}^{N} kt^{k-1} = \frac{d}{dt}\frac{t^{N+1}-t}{t-1}. $$

Multiplying the above equation by $t$ yields

$$ \sum_{k=1}^{N} kt^{k} = t\frac{d}{dt}\frac{t^{N+1}-t}{t-1}. $$

It is your job to do the differentiation and then substitute $t=a$.

science
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  • Does that identity help with $\sum\limits_{k=1}^Nka^k$? – robjohn Feb 23 '15 at 06:22
  • @robjohn: Yes it does! – science Feb 23 '15 at 06:29
  • Note that the tags do not include calculus, so I would not assume differentiation is available as a tool. – robjohn Feb 23 '15 at 06:34
  • @robjohn: The OP did not specify a specific technique! He just asked for summing the series and I gave him a technique to do that! – science Feb 23 '15 at 06:37
  • @science-can you please provide me a detailed explanation for $\sum\limits_{k=1}^Nka^k$

    I think its something like d/dx{f(x)}=$\sum\limits_{k=0}^{N-1}a^k$ but later I need to integrate it again and there will be some constant C to determine

     – user4131265 Feb 23 '15 at 06:50
0

Read all sums as sums from $0$ to $n$

Using the identity $\sum_k a^k=\frac{1-a^{N+1}}{1-a}$ we can calculate $\sum_k ka^k$

$\sum_k (k+1)a^k=1+a+a^2+...+a^N+a+a^2+....+a^N+a^2+a^3+...+a^N+...+a^N=\frac{1-a^{N+1}}{1-a}+a\frac{1-a^{N}}{1-a}+a^2\frac{1-a^{N-1}}{1-a}+....a^N\frac{1-a}{1-a}=\frac{1+a+a^2+...+a^N - Na^{N+1}}{1-a}=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}}{1-a}$

So $\sum_k ka^k=\sum_k (k+1)a^k-\sum_k a^k=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}-(1-a^{N+1})}{1-a}=\frac{1-Na^{N+1}+(N-1)a^{N+2}}{(1-a)^2}-\frac{1}{1-a}$

Leader47
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