Sum: $sum_{k=1}^{infty}{2k^2}/{5^{k}}$ [duplicate]
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This question already has an answer here:
How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
6 answers
$$sum_{k=1}^{infty}frac{2k^2}{5^k}$$
I came across this problem and I didn't get how to solve it at all. May someone explain this to me?
sequences-and-series summation
edited Dec 1 '18 at 17:03
amWhy
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
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marked as duplicate by Hans Lundmark, amWhy, Lord Shark the Unknown, Rebellos, Jean-Claude Arbaut Dec 1 '18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
6 answers
$$sum_{k=1}^{infty}frac{2k^2}{5^k}$$
I came across this problem and I didn't get how to solve it at all. May someone explain this to me?
sequences-and-series summation
edited Dec 1 '18 at 17:03
amWhy
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
$endgroup$
marked as duplicate by Hans Lundmark, amWhy, Lord Shark the Unknown, Rebellos, Jean-Claude Arbaut Dec 1 '18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
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– Jack D'Aurizio
Dec 1 '18 at 17:05
add a comment |
$begingroup$
This question already has an answer here:
How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
6 answers
$$sum_{k=1}^{infty}frac{2k^2}{5^k}$$
I came across this problem and I didn't get how to solve it at all. May someone explain this to me?
sequences-and-series summation
edited Dec 1 '18 at 17:03
amWhy
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
$endgroup$
This question already has an answer here:
How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
6 answers
$$sum_{k=1}^{infty}frac{2k^2}{5^k}$$
I came across this problem and I didn't get how to solve it at all. May someone explain this to me?
This question already has an answer here:
How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
6 answers
sequences-and-series summation
sequences-and-series summation
edited Dec 1 '18 at 17:03
amWhy
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
edited Dec 1 '18 at 17:03
amWhy
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
edited Dec 1 '18 at 17:03
amWhy
192k28225439
edited Dec 1 '18 at 17:03
amWhy
192k28225439
edited Dec 1 '18 at 17:03
amWhy
192k28225439
192k28225439
asked Dec 1 '18 at 16:57
weareallinweareallin
51
asked Dec 1 '18 at 16:57
weareallinweareallin
51
asked Dec 1 '18 at 16:57
weareallinweareallin
51
51
marked as duplicate by Hans Lundmark, amWhy, Lord Shark the Unknown, Rebellos, Jean-Claude Arbaut Dec 1 '18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Hans Lundmark, amWhy, Lord Shark the Unknown, Rebellos, Jean-Claude Arbaut Dec 1 '18 at 22:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
$endgroup$
– Jack D'Aurizio
Dec 1 '18 at 17:05
add a comment |
$begingroup$
$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
$endgroup$
– Jack D'Aurizio
Dec 1 '18 at 17:05
$begingroup$
$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
$endgroup$
– Jack D'Aurizio
Dec 1 '18 at 17:05
$begingroup$
$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
$endgroup$
– Jack D'Aurizio
Dec 1 '18 at 17:05
add a comment |
2 Answers
2
active
oldest
votes
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Hint: Write your sum as $$sum_{i=1}^{infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.
(Further hint: Start with the well-known power series for $frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
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Let $S=sum_{kgeq 1}frac{2k^2}{5^k}$. Then
$$begin{eqnarray*} 4S=5S-S=sum_{kgeq 1}frac{2k^2}{5^{k-1}}-sum_{kgeq 1}frac{2k^2}{5^k}&=&sum_{kgeq 0}frac{2(k+1)^2}{5^k}-sum_{kgeq 1}frac{2k^2}{5^k}\&=&2+sum_{kgeq 1}frac{2(2k+1)}{5^k}end{eqnarray*}$$
hence $S=frac{1}{2}+frac{1}{2}sum_{kgeq 1}frac{2k+1}{5^k}=frac{1}{2}+frac{T}{2}$. Now you may apply the same trick to $T$:
$$ 4T=5T-T = sum_{kgeq 0}frac{2k+3}{5^k}-sum_{kgeq 1}frac{2k+1}{5^k} = 3+sum_{kgeq 1}frac{2}{5^k} = 3+frac{frac{2}{5}}{1-frac{1}{5}}=frac{7}{2} $$
hence $T=frac{7}{8}$ and $S=frac{15}{16}$.
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Write your sum as $$sum_{i=1}^{infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.
(Further hint: Start with the well-known power series for $frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
$endgroup$
add a comment |
$begingroup$
Hint: Write your sum as $$sum_{i=1}^{infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.
(Further hint: Start with the well-known power series for $frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
$endgroup$
add a comment |
$begingroup$
Hint: Write your sum as $$sum_{i=1}^{infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.
(Further hint: Start with the well-known power series for $frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
$endgroup$
Hint: Write your sum as $$sum_{i=1}^{infty} 2k^2x^k $$ with $x=1/5$. That's a power series, so you only need to find which functions has has coefficients $2k^2$ in its power series.
(Further hint: Start with the well-known power series for $frac{1}{1-x}$. Differentiate it once, multiply by $x$ and differentiate it once again).
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
answered Dec 1 '18 at 17:03
Henning MakholmHenning Makholm
239k16303540
239k16303540
add a comment |
add a comment |
$begingroup$
Let $S=sum_{kgeq 1}frac{2k^2}{5^k}$. Then
$$begin{eqnarray*} 4S=5S-S=sum_{kgeq 1}frac{2k^2}{5^{k-1}}-sum_{kgeq 1}frac{2k^2}{5^k}&=&sum_{kgeq 0}frac{2(k+1)^2}{5^k}-sum_{kgeq 1}frac{2k^2}{5^k}\&=&2+sum_{kgeq 1}frac{2(2k+1)}{5^k}end{eqnarray*}$$
hence $S=frac{1}{2}+frac{1}{2}sum_{kgeq 1}frac{2k+1}{5^k}=frac{1}{2}+frac{T}{2}$. Now you may apply the same trick to $T$:
$$ 4T=5T-T = sum_{kgeq 0}frac{2k+3}{5^k}-sum_{kgeq 1}frac{2k+1}{5^k} = 3+sum_{kgeq 1}frac{2}{5^k} = 3+frac{frac{2}{5}}{1-frac{1}{5}}=frac{7}{2} $$
hence $T=frac{7}{8}$ and $S=frac{15}{16}$.
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let $S=sum_{kgeq 1}frac{2k^2}{5^k}$. Then
$$begin{eqnarray*} 4S=5S-S=sum_{kgeq 1}frac{2k^2}{5^{k-1}}-sum_{kgeq 1}frac{2k^2}{5^k}&=&sum_{kgeq 0}frac{2(k+1)^2}{5^k}-sum_{kgeq 1}frac{2k^2}{5^k}\&=&2+sum_{kgeq 1}frac{2(2k+1)}{5^k}end{eqnarray*}$$
hence $S=frac{1}{2}+frac{1}{2}sum_{kgeq 1}frac{2k+1}{5^k}=frac{1}{2}+frac{T}{2}$. Now you may apply the same trick to $T$:
$$ 4T=5T-T = sum_{kgeq 0}frac{2k+3}{5^k}-sum_{kgeq 1}frac{2k+1}{5^k} = 3+sum_{kgeq 1}frac{2}{5^k} = 3+frac{frac{2}{5}}{1-frac{1}{5}}=frac{7}{2} $$
hence $T=frac{7}{8}$ and $S=frac{15}{16}$.
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let $S=sum_{kgeq 1}frac{2k^2}{5^k}$. Then
$$begin{eqnarray*} 4S=5S-S=sum_{kgeq 1}frac{2k^2}{5^{k-1}}-sum_{kgeq 1}frac{2k^2}{5^k}&=&sum_{kgeq 0}frac{2(k+1)^2}{5^k}-sum_{kgeq 1}frac{2k^2}{5^k}\&=&2+sum_{kgeq 1}frac{2(2k+1)}{5^k}end{eqnarray*}$$
hence $S=frac{1}{2}+frac{1}{2}sum_{kgeq 1}frac{2k+1}{5^k}=frac{1}{2}+frac{T}{2}$. Now you may apply the same trick to $T$:
$$ 4T=5T-T = sum_{kgeq 0}frac{2k+3}{5^k}-sum_{kgeq 1}frac{2k+1}{5^k} = 3+sum_{kgeq 1}frac{2}{5^k} = 3+frac{frac{2}{5}}{1-frac{1}{5}}=frac{7}{2} $$
hence $T=frac{7}{8}$ and $S=frac{15}{16}$.
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
Let $S=sum_{kgeq 1}frac{2k^2}{5^k}$. Then
$$begin{eqnarray*} 4S=5S-S=sum_{kgeq 1}frac{2k^2}{5^{k-1}}-sum_{kgeq 1}frac{2k^2}{5^k}&=&sum_{kgeq 0}frac{2(k+1)^2}{5^k}-sum_{kgeq 1}frac{2k^2}{5^k}\&=&2+sum_{kgeq 1}frac{2(2k+1)}{5^k}end{eqnarray*}$$
hence $S=frac{1}{2}+frac{1}{2}sum_{kgeq 1}frac{2k+1}{5^k}=frac{1}{2}+frac{T}{2}$. Now you may apply the same trick to $T$:
$$ 4T=5T-T = sum_{kgeq 0}frac{2k+3}{5^k}-sum_{kgeq 1}frac{2k+1}{5^k} = 3+sum_{kgeq 1}frac{2}{5^k} = 3+frac{frac{2}{5}}{1-frac{1}{5}}=frac{7}{2} $$
hence $T=frac{7}{8}$ and $S=frac{15}{16}$.
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
$begingroup$
$k^2=2binom{k}{2}+binom{k}{1}$, then apply stars and bars.
$endgroup$
– Jack D'Aurizio
Dec 1 '18 at 17:05