Adding $n-1$ first terms of a sequence











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1
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Let's define a real - valued sequence:
$$a_k = kp^k,$$
where $p in mathbb{R}$ and $p in (0,1]$.

I would like to find the sum of first $n-1$ terms of $a_k$, that is
$$sum_{k=1}^{n-1} kp^k.$$
How can I do it?






real-analysis sequences-and-series







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  • 2




    Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
    – AnyAD
    Nov 20 at 12:39















up vote
1
down vote

favorite












Let's define a real - valued sequence:
$$a_k = kp^k,$$
where $p in mathbb{R}$ and $p in (0,1]$.

I would like to find the sum of first $n-1$ terms of $a_k$, that is
$$sum_{k=1}^{n-1} kp^k.$$
How can I do it?






real-analysis sequences-and-series







share|cite|improve this question


















  • 2




    Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
    – AnyAD
    Nov 20 at 12:39













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let's define a real - valued sequence:
$$a_k = kp^k,$$
where $p in mathbb{R}$ and $p in (0,1]$.

I would like to find the sum of first $n-1$ terms of $a_k$, that is
$$sum_{k=1}^{n-1} kp^k.$$
How can I do it?






real-analysis sequences-and-series







share|cite|improve this question













Let's define a real - valued sequence:
$$a_k = kp^k,$$
where $p in mathbb{R}$ and $p in (0,1]$.

I would like to find the sum of first $n-1$ terms of $a_k$, that is
$$sum_{k=1}^{n-1} kp^k.$$
How can I do it?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 12:30









Hendrra

1,025416




1,025416








  • 2




    Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
    – AnyAD
    Nov 20 at 12:39














  • 2




    Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
    – AnyAD
    Nov 20 at 12:39








2




2




Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
– AnyAD
Nov 20 at 12:39




Let $S $ denote the sum you want. Consider $S-Sp $ and use geometric sum formula.
– AnyAD
Nov 20 at 12:39










2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










Hint:



You probably know that $$sum_{k=0}^{n-1} x^k=frac{1-x^n}{1-x}tag{1}$$ Can you manipulate this sum to get $$sum_{k=1}^{n-1} kx^k$$ ? Try playing around with differentiation.




Edit for completion:



Differentiating both sides of $(1)$ yields $$sum_{k=1}^{n-1}kx^{k-1}=frac{(n-1)x^{n+1}-nx^n+x}{x(1-x)^2}tag{2}$$ (the series starts at $k=1$, because the first term is $0$) multiplying $(2)$ by $x$ yields the desired sum $$sum_{k=1}^{n-1}kx^{k}=frac{(n-1)x^{n+1}-nx^n+x}{(1-x)^2}tag{3}$$






share|cite|improve this answer























  • Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
    – Hendrra
    Nov 20 at 13:28


















up vote
1
down vote













Well, this is what you use if you don't know anything about differentiation -_-




Let $$S=p+2p^2+3p^3+cdots+(n-1)p^{n-1}.$$



Then $$Sp=p^2+2p^3+cdots+(n-2)p^{n-1}+(n-1)p^{n}$$



By subtracting the 1st equation with the 2nd equation we get



$$S(1-p)=p+p^2+p^3+cdots+p^{n-1}-(n-1)p^n=frac{1-p^n}{1-p}-1-(n-1)p^n$$






share|cite|improve this answer























  • The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
    – cansomeonehelpmeout
    Nov 26 at 10:25










  • Oh yeah thank you
    – Vee Hua Zhi
    Nov 27 at 4:42











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Hint:



You probably know that $$sum_{k=0}^{n-1} x^k=frac{1-x^n}{1-x}tag{1}$$ Can you manipulate this sum to get $$sum_{k=1}^{n-1} kx^k$$ ? Try playing around with differentiation.




Edit for completion:



Differentiating both sides of $(1)$ yields $$sum_{k=1}^{n-1}kx^{k-1}=frac{(n-1)x^{n+1}-nx^n+x}{x(1-x)^2}tag{2}$$ (the series starts at $k=1$, because the first term is $0$) multiplying $(2)$ by $x$ yields the desired sum $$sum_{k=1}^{n-1}kx^{k}=frac{(n-1)x^{n+1}-nx^n+x}{(1-x)^2}tag{3}$$






share|cite|improve this answer























  • Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
    – Hendrra
    Nov 20 at 13:28















up vote
5
down vote



accepted










Hint:



You probably know that $$sum_{k=0}^{n-1} x^k=frac{1-x^n}{1-x}tag{1}$$ Can you manipulate this sum to get $$sum_{k=1}^{n-1} kx^k$$ ? Try playing around with differentiation.




Edit for completion:



Differentiating both sides of $(1)$ yields $$sum_{k=1}^{n-1}kx^{k-1}=frac{(n-1)x^{n+1}-nx^n+x}{x(1-x)^2}tag{2}$$ (the series starts at $k=1$, because the first term is $0$) multiplying $(2)$ by $x$ yields the desired sum $$sum_{k=1}^{n-1}kx^{k}=frac{(n-1)x^{n+1}-nx^n+x}{(1-x)^2}tag{3}$$






share|cite|improve this answer























  • Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
    – Hendrra
    Nov 20 at 13:28













up vote
5
down vote



accepted







up vote
5
down vote



accepted






Hint:



You probably know that $$sum_{k=0}^{n-1} x^k=frac{1-x^n}{1-x}tag{1}$$ Can you manipulate this sum to get $$sum_{k=1}^{n-1} kx^k$$ ? Try playing around with differentiation.




Edit for completion:



Differentiating both sides of $(1)$ yields $$sum_{k=1}^{n-1}kx^{k-1}=frac{(n-1)x^{n+1}-nx^n+x}{x(1-x)^2}tag{2}$$ (the series starts at $k=1$, because the first term is $0$) multiplying $(2)$ by $x$ yields the desired sum $$sum_{k=1}^{n-1}kx^{k}=frac{(n-1)x^{n+1}-nx^n+x}{(1-x)^2}tag{3}$$






share|cite|improve this answer














Hint:



You probably know that $$sum_{k=0}^{n-1} x^k=frac{1-x^n}{1-x}tag{1}$$ Can you manipulate this sum to get $$sum_{k=1}^{n-1} kx^k$$ ? Try playing around with differentiation.




Edit for completion:



Differentiating both sides of $(1)$ yields $$sum_{k=1}^{n-1}kx^{k-1}=frac{(n-1)x^{n+1}-nx^n+x}{x(1-x)^2}tag{2}$$ (the series starts at $k=1$, because the first term is $0$) multiplying $(2)$ by $x$ yields the desired sum $$sum_{k=1}^{n-1}kx^{k}=frac{(n-1)x^{n+1}-nx^n+x}{(1-x)^2}tag{3}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 16:50

























answered Nov 20 at 12:34









cansomeonehelpmeout

6,5313834




6,5313834












  • Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
    – Hendrra
    Nov 20 at 13:28


















  • Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
    – Hendrra
    Nov 20 at 13:28
















Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
– Hendrra
Nov 20 at 13:28




Thank you! I knew that I can use differentiation is series but I didn't realize that finite sums are much more simpler and the same technique works :)
– Hendrra
Nov 20 at 13:28










up vote
1
down vote













Well, this is what you use if you don't know anything about differentiation -_-




Let $$S=p+2p^2+3p^3+cdots+(n-1)p^{n-1}.$$



Then $$Sp=p^2+2p^3+cdots+(n-2)p^{n-1}+(n-1)p^{n}$$



By subtracting the 1st equation with the 2nd equation we get



$$S(1-p)=p+p^2+p^3+cdots+p^{n-1}-(n-1)p^n=frac{1-p^n}{1-p}-1-(n-1)p^n$$






share|cite|improve this answer























  • The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
    – cansomeonehelpmeout
    Nov 26 at 10:25










  • Oh yeah thank you
    – Vee Hua Zhi
    Nov 27 at 4:42















up vote
1
down vote













Well, this is what you use if you don't know anything about differentiation -_-




Let $$S=p+2p^2+3p^3+cdots+(n-1)p^{n-1}.$$



Then $$Sp=p^2+2p^3+cdots+(n-2)p^{n-1}+(n-1)p^{n}$$



By subtracting the 1st equation with the 2nd equation we get



$$S(1-p)=p+p^2+p^3+cdots+p^{n-1}-(n-1)p^n=frac{1-p^n}{1-p}-1-(n-1)p^n$$






share|cite|improve this answer























  • The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
    – cansomeonehelpmeout
    Nov 26 at 10:25










  • Oh yeah thank you
    – Vee Hua Zhi
    Nov 27 at 4:42













up vote
1
down vote










up vote
1
down vote









Well, this is what you use if you don't know anything about differentiation -_-




Let $$S=p+2p^2+3p^3+cdots+(n-1)p^{n-1}.$$



Then $$Sp=p^2+2p^3+cdots+(n-2)p^{n-1}+(n-1)p^{n}$$



By subtracting the 1st equation with the 2nd equation we get



$$S(1-p)=p+p^2+p^3+cdots+p^{n-1}-(n-1)p^n=frac{1-p^n}{1-p}-1-(n-1)p^n$$






share|cite|improve this answer














Well, this is what you use if you don't know anything about differentiation -_-




Let $$S=p+2p^2+3p^3+cdots+(n-1)p^{n-1}.$$



Then $$Sp=p^2+2p^3+cdots+(n-2)p^{n-1}+(n-1)p^{n}$$



By subtracting the 1st equation with the 2nd equation we get



$$S(1-p)=p+p^2+p^3+cdots+p^{n-1}-(n-1)p^n=frac{1-p^n}{1-p}-1-(n-1)p^n$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 4:43

























answered Nov 20 at 12:45









Vee Hua Zhi

772124




772124












  • The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
    – cansomeonehelpmeout
    Nov 26 at 10:25










  • Oh yeah thank you
    – Vee Hua Zhi
    Nov 27 at 4:42


















  • The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
    – cansomeonehelpmeout
    Nov 26 at 10:25










  • Oh yeah thank you
    – Vee Hua Zhi
    Nov 27 at 4:42
















The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
– cansomeonehelpmeout
Nov 26 at 10:25




The expression for $S(1-p)$ is not right, it should be $$frac{1-p^n}{1-p}-1-(n-1)p^n$$ You need to subtract $1$, since $$frac{1-p^n}{1-p}=color{red}{1}+p+p^2+ldots+p^{n-1}$$
– cansomeonehelpmeout
Nov 26 at 10:25












Oh yeah thank you
– Vee Hua Zhi
Nov 27 at 4:42




Oh yeah thank you
– Vee Hua Zhi
Nov 27 at 4:42


















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