the real mathematical induction question....?

Question

The entire problem is like this,,,

[k(k+1)]^2 + (k+1)^3
------------
2

And the solution I have to get is
[k+1(k+1+1)]^2
-------------------
2

i'm trying to work backwards with the solution and forwards from the question in order to solve this. But in the solution part I dont know If I multiply everything inside and then multiply to the power of 2 or start from the power of 2 and then solving with what I got..
Thank you in advance for any help. :)

Thank you sooo much Pascal!!!!
yours is the best answer, but unfortunately i have to wait 4 hours to post it :)
Thank you very much.

Answer 1

Is this for the sum of the first n cubes formula? I can tell you what your problem is right now: that two should be _inside_ the brackets. Otherwise, this formula doesn't work.

Now, using the correct formula:
(k(k+1)/2)² + (k+1)³
We start by actually multiplying out the stuff in the parentheses:
((k²+k)/2)² + (k+1)³
(k^4 + 2k³ + k²)/4 + (k³ + 3k² + 3k + 1)
Now we give these fractions a common denominator:
(k^4 + 2k³ + k²)/4 + (4k³ + 12k² + 12k + 4)/4
And add:
(k^4 + 6k³ + 13k² + 12k + 4)/4
Factoring the polynomial:
(k+1)²(k+2)²/4
((k+1)(k+2)/2)²
Thus completing the proof.

Answer 2

The answer I and Steven Hawkins arrived at is

[mc+ cm]+^0220{745}74/866%=47>9835+xrc|..124 or to the non mathmetical among you B/0>LL{0}^c¬¬K><>

Answer 3

to prove thesum of the cubes of 'n' natutal numbers is
=[n(n+1)/2]^2
true for n=1
let it be true for n=k
Sk=[k(k+1)/2]^2
S(k+1)=1#+2^3+3^3+.......+k^3+(k+1)^3
=[k(k+1)/2]^2+(k+1)^3
=[k(k+1)/2]^2]+[4(k+1)^3/4]
taking out [(k+1)/2]^2 common
=[(k+1)/2]^2[k^2+4k+4]
=([(k+1)/2]^2(k+2)^2
=[(k+1)(k+2)/2]^2
so true for n=k+1 also
so true for any 'n'
hence proved

Answer 4

there is no way at the instant TO tutor THAT GOD IS genuine OR UNREAL, IF IT have been THEN THERE might basically BE a million faith that all and sundry might stick to, people ought to GET THIS by THEIR HEAD.