Mixing
Mathematical induction find values of skipping numbers.
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I have a question which is finding the value of a skipping numbers mathematical induction problem.
The problem is as follows:
Prove by mathematical induction, that $$1^3+2^3+3^3+dots+n^3=frac{n^2(n+1)^2}{4}.$$
I can successfully prove this.
The next part asks to find the value of
$$2^3+4^3+6^3+dots+30^3$$
I can find the sum of up to thirty but I don't know what to subtract it from. Thank you for your help
algebra-precalculus
edited Jan 23 at 8:06
Robert Z
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
$endgroup$
add a comment |
$begingroup$
I have a question which is finding the value of a skipping numbers mathematical induction problem.
The problem is as follows:
Prove by mathematical induction, that $$1^3+2^3+3^3+dots+n^3=frac{n^2(n+1)^2}{4}.$$
I can successfully prove this.
The next part asks to find the value of
$$2^3+4^3+6^3+dots+30^3$$
I can find the sum of up to thirty but I don't know what to subtract it from. Thank you for your help
algebra-precalculus
edited Jan 23 at 8:06
Robert Z
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
$endgroup$
2
$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18
add a comment |
$begingroup$
I have a question which is finding the value of a skipping numbers mathematical induction problem.
The problem is as follows:
Prove by mathematical induction, that $$1^3+2^3+3^3+dots+n^3=frac{n^2(n+1)^2}{4}.$$
I can successfully prove this.
The next part asks to find the value of
$$2^3+4^3+6^3+dots+30^3$$
I can find the sum of up to thirty but I don't know what to subtract it from. Thank you for your help
algebra-precalculus
edited Jan 23 at 8:06
Robert Z
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
$endgroup$
I have a question which is finding the value of a skipping numbers mathematical induction problem.
The problem is as follows:
Prove by mathematical induction, that $$1^3+2^3+3^3+dots+n^3=frac{n^2(n+1)^2}{4}.$$
I can successfully prove this.
The next part asks to find the value of
$$2^3+4^3+6^3+dots+30^3$$
I can find the sum of up to thirty but I don't know what to subtract it from. Thank you for your help
algebra-precalculus
algebra-precalculus
edited Jan 23 at 8:06
Robert Z
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
edited Jan 23 at 8:06
Robert Z
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
edited Jan 23 at 8:06
Robert Z
100k1069140
edited Jan 23 at 8:06
Robert Z
100k1069140
edited Jan 23 at 8:06
Robert Z
100k1069140
100k1069140
asked Jan 23 at 7:55
Horace LauHorace Lau
31
asked Jan 23 at 7:55
Horace LauHorace Lau
31
asked Jan 23 at 7:55
Horace LauHorace Lau
31
31
2
$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18
add a comment |
2
$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18
2
2
$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Simply observe that $$begin{align}
2^3+4^3+6^3+cdots+30^3&=(2times1)^3+(2times2)^3+(2times3)^3+cdots+(2times15)^3\
&=2^3left(1^3+2^3+3^3+cdots15^3right)\
&=2^3left[frac{15^2(15+1)^2}{4}right]\
&=115200
end{align}$$
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
$endgroup$
add a comment |
$begingroup$
What happens if you divide the sum by $2^3$? :)
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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oldest
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$begingroup$
Simply observe that $$begin{align}
2^3+4^3+6^3+cdots+30^3&=(2times1)^3+(2times2)^3+(2times3)^3+cdots+(2times15)^3\
&=2^3left(1^3+2^3+3^3+cdots15^3right)\
&=2^3left[frac{15^2(15+1)^2}{4}right]\
&=115200
end{align}$$
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
$endgroup$
add a comment |
$begingroup$
Simply observe that $$begin{align}
2^3+4^3+6^3+cdots+30^3&=(2times1)^3+(2times2)^3+(2times3)^3+cdots+(2times15)^3\
&=2^3left(1^3+2^3+3^3+cdots15^3right)\
&=2^3left[frac{15^2(15+1)^2}{4}right]\
&=115200
end{align}$$
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
$endgroup$
add a comment |
$begingroup$
Simply observe that $$begin{align}
2^3+4^3+6^3+cdots+30^3&=(2times1)^3+(2times2)^3+(2times3)^3+cdots+(2times15)^3\
&=2^3left(1^3+2^3+3^3+cdots15^3right)\
&=2^3left[frac{15^2(15+1)^2}{4}right]\
&=115200
end{align}$$
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
$endgroup$
Simply observe that $$begin{align}
2^3+4^3+6^3+cdots+30^3&=(2times1)^3+(2times2)^3+(2times3)^3+cdots+(2times15)^3\
&=2^3left(1^3+2^3+3^3+cdots15^3right)\
&=2^3left[frac{15^2(15+1)^2}{4}right]\
&=115200
end{align}$$
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
answered Jan 23 at 8:12
Faiq IrfanFaiq Irfan
826317
826317
add a comment |
add a comment |
$begingroup$
What happens if you divide the sum by $2^3$? :)
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
$endgroup$
add a comment |
$begingroup$
What happens if you divide the sum by $2^3$? :)
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
$endgroup$
add a comment |
$begingroup$
What happens if you divide the sum by $2^3$? :)
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
$endgroup$
What happens if you divide the sum by $2^3$? :)
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
answered Jan 23 at 8:00
angryavianangryavian
42.1k23381
42.1k23381
add a comment |
add a comment |
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$begingroup$
How about set $n=15$ and multiply both sides by $2^3=8$...
$endgroup$
– TheSimpliFire
Jan 23 at 8:08
$begingroup$
Thank you @TheSimpliFire
$endgroup$
– Horace Lau
Jan 23 at 8:18