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# Sum of Cubes - Odd Integers

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Show  by induction  that  the sum of the cubes of the first positive integers is ¼n2 (n + 1)2 and deduce that  the  sum  of  the  cubes  of  the  n + odd integers  from 1 to (2n  + 1) inclusive  is:

(n + 1)2 (2n2   + 4n + 1).

OldTimer  Feb 26, 2018
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#1
+83946
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Here's the first part :

1^3   + 2^3  +  3^3   +  .....+  n^3   =  n^2 (n + 1)^2  / 4

Show that it's true for  n  = 1

1^3   =   (1)^2 (1 + 1)^2 / 4   =  4 / 4  =  1   ⇒   true

Assume that it;s true for n  = k

So

1^3 + 2^3  + 3^3  +  .... +   k^3   =   k^2 (k + 1)^2 / 4

Show that it's true for k + 1

That is

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  [k+1]^2 [ (k +1) + 1) ]^2 / 4

So we have

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  k^2 ( k + 1)^2 / 4  + (k + 1)^3

= [  k^2 (k + 1)^2  + 4(k + 1)^3] / 4

=  [ (k + 1)^2 ( k^2 + 4(k + 1) ]  / 4

= [ (k + 1)^2 ( k^2 + 4k + 4) ] /4

= [ (k + 1)^2 ( k + 2)^2 / 4 =

= [ (k + 1)^2 [ (k + 1) + 1 ]^2 / 4

Which is what we wanted to show

For the second part....we can evaluate these sum-of-diferences  [ although there are probably faster ways, maybe  ??? ]

1         28         153          496          1225              2556

27         125         343         729              1331

98           218       386           602

120          168       216

48           48

We will generate a 4th power polynomial.....which is consistent with your second formula

Sorry... got to go to bed, now  !!!   ....LOL!!!

CPhill  Feb 26, 2018
#2
+42
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HI Cphill

I did not really understand exactly second part of question but I have interpreted it as follows:

Define Pattern :

Sc = 13 + 33 + 53..........+ (2n-1)3  =  n2(2n2-1)  (as per formula for sum of cubes for odd integers)

Now Sum of cubes for n+(n+1)

S2n+1 = n2(2n2-1) + [ 2(n+1)-1]3

= 2n4- n2 + [2n+1]3

= 2n4- n2  + 8n3+ 12n+ 6n + 1

= (2n4 + 4n3 + 2n2) + (4n3 + 8n2 + 4n) + (n2 + 2n + 1)

= 2n2(n+1)2 + 4n(n+1)2 + (n+1)2

(n+1)2(2n2 + 4n + 1)

OldTimer  Mar 4, 2018

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