# Math Help - [SOLVED] Need help with Mathematical Induction problem.

1. ## [SOLVED] Need help with Mathematical Induction problem.

Hello! Could someone here please help me with the following problem? My lecture notes are pretty vague with the explainations and to mathematical inductions and no examples were given. My teacher is not clear with the explainations too!Thank you in advance!

1) Using a suitable form of Mathematical Induction prove the following for all
integer n ≥ 1.

$
\sum_{j = 1}^nj^3 = \frac{{n}^2({n}+{1})^2}{4}
$

2. -First of all "j" and "f" are same

-To prove it show that this statement it is correct for j=1

-Next step; Consider that this statement is true for j=k
(where k is any arbitrary constant)

-Using the above condition prove this statement is true for j=k+1

EDIT :This proves the statement for any term , Please tell incase you find difficulty in implementing the method

3. Let CLAIM(n) be $
\sum_{j = 1}^nj^3 = \frac{{n}^2({n}+{1})^2}{4}
$

Step 1:
CLAIM(1) is $
\sum_{j = 1}^1j^3 = \frac{{1}^2({1}+{1})^2}{4}
$

LHS = 1 ; RHS = 1
Therefore, LHS=RHS, So CLAIM(1) is true.

CLAIM(2) is $
\sum_{j = 1}^2j^3 = \frac{{2}^2({2}+{1})^2}{4}
$

LHS = 9 ; RHS = 9
Therefore, LHS=RHS, So CLAIM(2) is true.

Are my initial steps above correct?

If it is so, i'm still pretty stucked at what to do in the following steps, like how do i actually go about implementing (K+1) into the equation? Please guide me along ADARSH! Thanks!

4. Originally Posted by Soap
Let CLAIM(n) be $
\sum_{j = 1}^nj^3 = \frac{{n}^2({n}+{1})^2}{4}
$

Step 1:
CLAIM(1) is $
\sum_{j = 1}^1j^3 = \frac{{1}^2({1}+{1})^2}{4}
$

LHS = 1 ; RHS = 1
Therefore, LHS=RHS, So CLAIM(1) is true.

Assumption(k) is $
\sum_{j = 1}^kj^3 = \frac{{k}^2({k}+{1})^2}{4}
$

CLAIM(k+1) : Now do it using the assumption we made

. The faster way is Watch MrF

5. Originally Posted by Soap
Let CLAIM(n) be $
\sum_{j = 1}^nj^3 = \frac{{n}^2({n}+{1})^2}{4}
$

Step 1:
CLAIM(1) is $
\sum_{j = 1}^1j^3 = \frac{{1}^2({1}+{1})^2}{4}
$

LHS = 1 ; RHS = 1
Therefore, LHS=RHS, So CLAIM(1) is true.

CLAIM(2) is $
\sum_{j = 1}^2j^3 = \frac{{2}^2({2}+{1})^2}{4}
$

LHS = 9 ; RHS = 9
Therefore, LHS=RHS, So CLAIM(2) is true.

Are my initial steps above correct?

If it is so, i'm still pretty stucked at what to do in the following steps, like how do i actually go about implementing (K+1) into the equation? Please guide me along ADARSH! Thanks!
Step 2 (inductive hypothesis): Assume $\sum_{j = 1}^k j^3 = \frac{{k}^2({k}+{1})^2}{4}$.

Step 3: Assuming step 2 is true, show that $\sum_{j = 1}^{k+1} j^3 = \frac{{(k+1)}^2({k}+{2})^2}{4}$.

To do this, you should start by noting that

$\sum_{j = 1}^{k+1} j^3 = \sum_{j = 1}^{k} j^3 + (k+1)^3$

$= \frac{{k}^2({k}+{1})^2}{4} + (k+1)^3$ using the inductive hypothesis (Step 2)

= ....

6. Step 2:
Assume CLAIM(k) is true,
$
\sum_{j = 1}^kj^3 = \frac{{k}^2({k}+{1})^2}{4}
$

Prove CLAIM(k+1) is true, that is, prove that
$
\sum_{j = 1}^{k+1}j^3 = \frac{{{k}+{1}}^2({{k}+{1}}+{1})^2}{4}
$

= $
\frac{{{k}+{1}}^2({k}+{2})^2}{4}
$

Are the steps right?

7. Originally Posted by Soap
Step 2:
Assume CLAIM(k) is true,
$
\sum_{j = 1}^kj^3 = \frac{{k}^2({k}+{1})^2}{4}
$

Prove CLAIM(k+1) is true, that is, prove that
$
\sum_{j = 1}^{k+1}j^3 = \frac{{{k}+{1}}^2({{k}+{1}}+{1})^2}{4}
$

= $
\frac{{{k}+{1}}^2({k}+{2})^2}{4}
$

Are the steps right?
Yes now prove it

Hint as MrF gave :You need to prove that

$
1^3 +2^3 .....(k)^3 +(k+1)^3 =
\frac{({k}+{1})^2({k}+{2})^2}{4}
$

You can use
$1^3+2^3.....k^3 = \frac{{k}^2({k}+{1})^2}{4}$

8. Originally Posted by Soap
Step 2:
Assume CLAIM(k) is true,
$
\sum_{j = 1}^kj^3 = \frac{{k}^2({k}+{1})^2}{4}
$

Prove CLAIM(k+1) is true, that is, prove that
$
\sum_{j = 1}^{k+1}j^3 = \frac{{\color{red}(}k+1{\color{red})}^2 (k+1+1)^2}{4}
$

= $
\frac{{\color{red}(} k+1{\color{red})}^2 (k+2)^2}{4}
$

Are the steps right?
I assume you meant to include the red brackets around the first term of the numerator.

Did you read my earlier reply? Then you should know the answer to that question. And you should know what to do next (since I showed you ....). All that's left is a small bit of algebra that has nothing to do with induction so

"My notes are pretty vague with the explainations and to mathematical inductions and no examples were given. My teacher is not clear with the explainations too"

is now irrelevant.

9. LHS = $\sum_{j = 1}^kj^3
$

= $
\sum_{j = 1}^kj^3 + ({k}+{1})^3
$

= $\frac{({k}+{1})^2({k}+{2})^2}{4} + ({k}+{1})^3$
= $\frac{k^2({k}+{1})^2 + {4}({k}+{1})^3}{4}$
= $\frac{k^2({k}+{1})({k}+{1}) + {4}({k}+{1})({k}+{1})({k}+{1})}{4}$
= $\frac{k^2({k^2}+{2k}+{1}) + {4}({k^2}+{2k}+{1})({k}+{1})}{4}$
= $\frac{({k^2}+{4})({k^2}+{2k}+{1})({k}+{1})}{4}$

Uhmm is there any mistakes in my steps? And how to prove that it is equal to $\frac{({k}+{1})^2({k}+{2})^2}{4}$ ?

10. Originally Posted by Soap
LHS = $\sum_{j = 1}^kj^3
$

= $
\sum_{j = 1}^kj^3 + ({k}+{1})^3
$

= $\frac{({k}+{1})^2({k}+{2})^2}{4} + ({k}+{1})^3$
= $\frac{k^2({k}+{1})^2 + {4}({k}+{1})^3}{4}$
= $\frac{k^2({k}+{1})({k}+{1}) + {4}({k}+{1})({k}+{1})({k}+{1})}{4}$
= $\frac{k^2({k^2}+{2k}+{1}) + {4}({k^2}+{2k}+{1})({k}+{1})}{4}$
= $\frac{({k^2}+{4})({k^2}+{2k}+{1})({k}+{1})}{4}$

Uhmm is there any mistakes in my steps? And how to prove that it is equal to $\frac{({k}+{1})^2({k}+{2})^2}{4}$
$k^2 (k+1)^2 + 4 (k+1)^3 = (k+1)^2 [k^2 + 4(k+1)]$. The next steps are (I should hope) obvious.

11. Originally Posted by mr fantastic
$k^2 (k+1)^2 + 4 (k+1)^3 = (k+1)^2 [k^2 + 4(k+1)]$. The next steps are (I should hope) obvious.
Hmmm.. but for

$k^2 (k+1)^2 + 4 (k+1)^{\color{red}3}{\color{red}} = (k+1)^2 [k^2 + 4(k+1)]$

What happens to the cube in red? Where did it disappeared to? If it is without the cube, i would understand the equation. My apologies, my math isn't good. So please do bear with me.

12. Originally Posted by Soap
Hmmm.. but for

$k^2 (k+1)^2 + 4 (k+1)^{\color{red}3}{\color{red}} = (k+1)^2 [k^2 + 4(k+1)]$

What happens to the cube in red? Where did it disappeared to? If it is without the cube, i would understand the equation. My apologies, my math isn't good. So please do bear with me.
See this
$k^2 (k+1)^2 + 4 (k+1)^{\color{red}3}{\color{red}}= (k+1)^2 [k^2 + 4(k+1)]$

Take (k+1)^2 as common you will be left with
$
(k+1)^2[k^2 + 4 (k+1)]/4
$

13. Ahhhh!!! Thanks a lot ADARSH and mr_fantastic! I finally understood how it should be done. Well guess practices makes perfect! Really glad i came here for help! Kudos to you both!