Missing in between terms during the process of mathematical induction
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I need someone's idea here... im trying to prove a pattern we formulated with my classmate as we deal with the even powers of natural numbers that can be express as a sum of consecutive odd numbers starting from 1 here's the formula we come up with$$sum_{i=1}^{n^{k}} (2i - 1) = n^{2k}$$
For k = 1,
$$sum_{i=1}^{n} (2i - 1) = n^{2}$$
Which can be easily proven by mathematical induction but when k = 2, it will become,$$sum_{i=1}^{n^{2}} (2i - 1) = n^{4}$$ and when im doing the mathematical induction i got stuck for some terms in between n and n + 1 such as
$1 + 3 + 5 + ...+ (2n^{k} - 1)$ + (some in between terms i cant express in summation) + $(2(n +1)^{k} -1) = (n +1)^{4}$
Actually i want to prove by mathematical induction the general k, but i struggle so much that i decided to manually assume values for k = 1,2,3... but it seems that i cant manage to deal with k = 2 . Any ideas? Thanks in advance!
number-theory
asked Jan 15 at 9:53
rosarosa
604616
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add a comment |
$begingroup$
I need someone's idea here... im trying to prove a pattern we formulated with my classmate as we deal with the even powers of natural numbers that can be express as a sum of consecutive odd numbers starting from 1 here's the formula we come up with$$sum_{i=1}^{n^{k}} (2i - 1) = n^{2k}$$
For k = 1,
$$sum_{i=1}^{n} (2i - 1) = n^{2}$$
Which can be easily proven by mathematical induction but when k = 2, it will become,$$sum_{i=1}^{n^{2}} (2i - 1) = n^{4}$$ and when im doing the mathematical induction i got stuck for some terms in between n and n + 1 such as
$1 + 3 + 5 + ...+ (2n^{k} - 1)$ + (some in between terms i cant express in summation) + $(2(n +1)^{k} -1) = (n +1)^{4}$
Actually i want to prove by mathematical induction the general k, but i struggle so much that i decided to manually assume values for k = 1,2,3... but it seems that i cant manage to deal with k = 2 . Any ideas? Thanks in advance!
number-theory
asked Jan 15 at 9:53
rosarosa
604616
$endgroup$
$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58
add a comment |
$begingroup$
I need someone's idea here... im trying to prove a pattern we formulated with my classmate as we deal with the even powers of natural numbers that can be express as a sum of consecutive odd numbers starting from 1 here's the formula we come up with$$sum_{i=1}^{n^{k}} (2i - 1) = n^{2k}$$
For k = 1,
$$sum_{i=1}^{n} (2i - 1) = n^{2}$$
Which can be easily proven by mathematical induction but when k = 2, it will become,$$sum_{i=1}^{n^{2}} (2i - 1) = n^{4}$$ and when im doing the mathematical induction i got stuck for some terms in between n and n + 1 such as
$1 + 3 + 5 + ...+ (2n^{k} - 1)$ + (some in between terms i cant express in summation) + $(2(n +1)^{k} -1) = (n +1)^{4}$
Actually i want to prove by mathematical induction the general k, but i struggle so much that i decided to manually assume values for k = 1,2,3... but it seems that i cant manage to deal with k = 2 . Any ideas? Thanks in advance!
number-theory
asked Jan 15 at 9:53
rosarosa
604616
$endgroup$
I need someone's idea here... im trying to prove a pattern we formulated with my classmate as we deal with the even powers of natural numbers that can be express as a sum of consecutive odd numbers starting from 1 here's the formula we come up with$$sum_{i=1}^{n^{k}} (2i - 1) = n^{2k}$$
For k = 1,
$$sum_{i=1}^{n} (2i - 1) = n^{2}$$
Which can be easily proven by mathematical induction but when k = 2, it will become,$$sum_{i=1}^{n^{2}} (2i - 1) = n^{4}$$ and when im doing the mathematical induction i got stuck for some terms in between n and n + 1 such as
$1 + 3 + 5 + ...+ (2n^{k} - 1)$ + (some in between terms i cant express in summation) + $(2(n +1)^{k} -1) = (n +1)^{4}$
Actually i want to prove by mathematical induction the general k, but i struggle so much that i decided to manually assume values for k = 1,2,3... but it seems that i cant manage to deal with k = 2 . Any ideas? Thanks in advance!
number-theory
number-theory
asked Jan 15 at 9:53
rosarosa
604616
asked Jan 15 at 9:53
rosarosa
604616
asked Jan 15 at 9:53
rosarosa
604616
asked Jan 15 at 9:53
rosarosa
604616
asked Jan 15 at 9:53
rosarosa
604616
604616
$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58
add a comment |
$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58
$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58
$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58
add a comment |
1 Answer
1
active
oldest
votes
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Your statement can easily be shown without induction, simply using the fact that
$$sum_{i=1}^N = frac{N(N+1)}{2}$$
and $$sum_{i=1}^N 1 = N.$$
This is because
$$sum_{i=1}^{n^k}(2i-1) = 2cdotsum_{i=1}^{n^k} i - sum_{i=1}^{n^k} 1.$$
answered Jan 15 at 9:59
5xum5xum
92.8k395162
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$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
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– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Your statement can easily be shown without induction, simply using the fact that
$$sum_{i=1}^N = frac{N(N+1)}{2}$$
and $$sum_{i=1}^N 1 = N.$$
This is because
$$sum_{i=1}^{n^k}(2i-1) = 2cdotsum_{i=1}^{n^k} i - sum_{i=1}^{n^k} 1.$$
answered Jan 15 at 9:59
5xum5xum
92.8k395162
$endgroup$
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
add a comment |
$begingroup$
Your statement can easily be shown without induction, simply using the fact that
$$sum_{i=1}^N = frac{N(N+1)}{2}$$
and $$sum_{i=1}^N 1 = N.$$
This is because
$$sum_{i=1}^{n^k}(2i-1) = 2cdotsum_{i=1}^{n^k} i - sum_{i=1}^{n^k} 1.$$
answered Jan 15 at 9:59
5xum5xum
92.8k395162
$endgroup$
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
add a comment |
$begingroup$
Your statement can easily be shown without induction, simply using the fact that
$$sum_{i=1}^N = frac{N(N+1)}{2}$$
and $$sum_{i=1}^N 1 = N.$$
This is because
$$sum_{i=1}^{n^k}(2i-1) = 2cdotsum_{i=1}^{n^k} i - sum_{i=1}^{n^k} 1.$$
answered Jan 15 at 9:59
5xum5xum
92.8k395162
$endgroup$
Your statement can easily be shown without induction, simply using the fact that
$$sum_{i=1}^N = frac{N(N+1)}{2}$$
and $$sum_{i=1}^N 1 = N.$$
This is because
$$sum_{i=1}^{n^k}(2i-1) = 2cdotsum_{i=1}^{n^k} i - sum_{i=1}^{n^k} 1.$$
answered Jan 15 at 9:59
5xum5xum
92.8k395162
answered Jan 15 at 9:59
5xum5xum
92.8k395162
answered Jan 15 at 9:59
5xum5xum
92.8k395162
answered Jan 15 at 9:59
5xum5xum
92.8k395162
92.8k395162
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
add a comment |
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
I get ur point sir thanks a lot, i thought there is a need for mathematical induction to prove that the equation will hold for every possible values...
$endgroup$
– rosa
Jan 15 at 10:50
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
@rosa Well, the expression $sum_{i=1}^N=frac{N(N+1)}{2}$ is usually proven through induction, so techincally, you are using induction.
$endgroup$
– 5xum
Jan 15 at 10:52
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
$begingroup$
Aw.. ur right.. i thought i must undergo the usual step in induction..thanks a lot
$endgroup$
– rosa
Jan 15 at 11:04
add a comment |
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$begingroup$
Why are you only considering sums to a power of $n$?
$endgroup$
– Tobias Kildetoft
Jan 15 at 9:58