Please help me!

If it's "≤"....then my first answer is the correct one......n = 7

We can prove, by induction, that  the sum of the first n positive integers squared = the sum of the cubes of the first n positive integers.

So n here, = 7

I think this is wrong, incorrect

If this is a democracy, I vote for CPhill's answer.

I have an online class, and it is wrong. Sorry

It might not be the largest integer

Sorry, I thought there was an equal sign, not a less than sign.

Ok, I have time. No worries.

So, what is the answer?

CPhill? Can you help me please?

OK, Anonymous, since you forced my hand...

Show that it is true for n = 2....that is

(1 + 2)^2 = 1^3 + 2^3 

3^2 = 1 + 8

9 = 9

Assume it is true for k = 1...that is ....

(1 + 2 + 3 +........+ (k-1) + k)^2 = 1^3 + 2^3 + 3^3 + ....+ (k-1)^3 + k^3

Show that it is true for k +1....so....

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

And by an identity, the left side can be written as

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

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

(1^3 + 2^3 + 3^3 +.....+(k-1)^3 +k^3) + 2( k+ 1)(1+2 +3 +......+k) + (k +1)^2

And setting this = to the right hand side and subtracting like terms from both sides, we have

2(k+1)(1+2+3+....+k) + (k+1)^2 = (k+1)^3      divide through by (k+1)

2(1+2+3+......+k) + (k+1) = (k + 1)^2   subtract (k+1) from each side

2(1 + 2 + 3 +.......+ k) = (k+1)^2 - (k+1)       factor the right side

2(1 + 2 + 3 +.......+ k) = (k+1)[(k+1)-1]

2(1 + 2 + 3 +.......+ k) = (k+1)[(k]    divide boh sides by 2

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

And the right side is just the "formula" for ths sum of the first k positive integers, i.e, the left side.......

So, what is the answer? Sorry, it is confusing.

Sorry .....I misread the '<" sign as an "="

I'll amend my answer to n= 6....!!!   My bad !!!

ok, thanks

=<

-Robert

not Thought about it, eh?