Question #81d37

2 Answers
Mar 16, 2015

This is a classical case in which you can prove something by induction. The idea is the following: think of a river with very strong current that you must cross. You can't swim, but there are several stones, and you can jump from one to the other. If you find a first (solid!) stone, and you learn how to jump from a stone to the next, then you can cross the river!

In formulas, a proof by induction works as follows:

  1. First of all, you must prove that the formula holds for a basic case (which is usually #n=0# or #n=1#)
  2. You must prove that, if the formula holds for a generic value #n#, then it also works for #n+1#.

By doing so, you can work -indeed- by induction, and step from a number to its successor, and so cover the entire range of natural numbers: the statement it's true for #n=1#. Then, it's true for #n=2#. Applying the same logic, we can step from #n=2# to #n=3#, and so on.

So, let's begin with the basic case #n=1#. Is it true that

#\sum_{i=1}^1 i^3= {1^2*(1+1)^2}/4#?

Of course it is, since the sum means simply 1^3, which is one, and the fraction on the right side is also 1.

So now, let's assume that the sum of the first #n# cubes is #{n^2*(n+1)^2}/4#, and we must prove that the sum of the first #n+1# cubes is #{(n+1)^2*[(n+1)+1]^2}/4#, which is the previous formula in which I substituted #n\rightarrow n+1#. We have:

#\sum_{i=1}^{n+1} i^3=\sum_{i=1}^{n} i^3 + (n+1)^3#, and we are assuming that the sum up to #n# equals #{n^2*(n+1)^2}/4#. We simply need to add #(n+1)^3# and check the result:

#{n^2*(n+1)^2}/4+(n+1)^3=#
#{n^2 (n^2+2n+1) + 4(n^3+3n^2+3n+1)}/4=#
#{n^4+2n^3+n^2+4n^3+12n^2+12n+4}/4=#
#{n^4+6n^3+13n^2+12n+4}/4#

We want this to be equal to #{(n+1)^2*[(n+1)+1]^2}/4#, which is true if and only if the numerators are equal. Let's expand this last numerator:
#(n+1)^2*[(n+1)+1]^2=#
#(n^2+2n+1)(n+2)^2=#
#(n^2+2n+1)(n^2+4n+4)=#
#n^4+4n^3+4n^2+2n^3+8n^2+8n+n^2+4n+4=#
#n^4+6n^3+13n^2+12n+4#

which is exactly what we wanted.

Mar 16, 2015

First of all let's try to understand why it works.

We know that #sum_(i=1)^ni=(n(n+1))/2#,

and this is because we have #n/2# sums of the value #n+1# (that is because the first added to the last, the second added to the second-last, and so on... have all value #n+1#).

Now we can see our sum:

#sum_(i=1)^ni^3=1+8+27+64+125+...+n^3#

Let's proceed step by step:

#S_1=sum_(i=1)^1i^3=1#,

#S_1^'=sum_(i=1)^1i=1#,

so #S_1= (S_1^')^2#.

#S_2=sum_(i=1)^2i^3=1+8=9#,

#S_2^'=sum_(i=1)^2i=1+2=3#,

so #S_2= (S_2^')^2#.

#S_3=sum_(i=1)^3i^3=1+8+27=36#,

#S_3^'=sum_(i=1)^3i=1+2+3=6#,

so #S_3= (S_3^')^2#.

...

#S_n=sum_(i=1)^ni^3=1+8+27+64+125+...+n^3#,

#S_n^'=sum_(i=1)^1i=1+2+3+4+5+...+n=(n(n+1))/2#,

so #S_n# has to be #(S_n^')^2# and so:

#S_n=((n(n+1))/2)^2=(n^2(n+1)^2)/4#.

But this is not a demonstration.

This is only an idea.

We have to demonstrate it with the induction rule, that says:

If a property is true for the first value, we can assume that it is true for the #n#th term and we have to demonstrate that it is true for the #(n+1)#th term.

So:

Is the property true for the first term?

#sum_(i=1)^1i^3=1# and #(1^2(1+1)^2)/4=1#: YES

Let's assume that it is true for #n#:

#sum_(i=1)^ni^3=(n^2(n+1)^2)/4# and this is our hypotesis.

Let's demonstrate that it is true also for #n+1#:

#sum_(i=1)^(n+1)i^3=sum_(i=1)^ni^3+(n+1)^3=#

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

#=((n+1)^2(n^2+4n+4))/4=((n+1)^2(n+2)^2)/4#

that is perfectly the formula that we have to demonstrate in which at the place of #n# there is #n+1#!