+0  
 
0
100
2
avatar+46 

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
Sort: 

2+0 Answers

 #1
avatar+86565 
+2

 

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!!!

 

 

 

cool cool cool

CPhill  Feb 26, 2018
 #2
avatar+46 
0

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

15 Online Users

avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy