SOLUTION: Use Mathematical Induction to show that the following statement is true for all natural numbers n: {{{ 1^3+2^3+3^3+...+}}} {{{n^3 = n^2(n+1)^2/(4) }}}
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-> SOLUTION: Use Mathematical Induction to show that the following statement is true for all natural numbers n: {{{ 1^3+2^3+3^3+...+}}} {{{n^3 = n^2(n+1)^2/(4) }}}
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You can put this solution on YOUR website! Use Mathematical Induction to show that the following statement is true for all natural numbers n: 1^3+2^3+3^3+...+ n^3 = n^2(n+1)^2/(4)
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Show it is true for n = 1::
1^3 = [1^2(1+1)^2]/4 = (1*2^2)/4
1 = 4/4
1 = 1
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Assume it is true for n = k,
1^3 + 2^3 + ... + k^3 = [k^2(k+1)^2]/4
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Show it is true for n = k+1
1^3 + 2^3 + ..+ k^3 + (k+1)^3 = [k^2(k+1)^2]/4 + (k+1)^3
Factor to getl::
= (k+1)^2[k^2/4 + (k+1)]
= (k+1)^2[(k^2 + 4k+ 4)]/4
= (k+1)^2[((k+1)+1)^2]/4
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So it is true for n = k+1.
Cheers,
Stan H.
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You can put this solution on YOUR website! For n=2:
and
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For n=3:
and
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Assume it is true for n=k. That is + ... + (*)
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Now let n=k+1: + ... +
Which, upon substituting (*), gets us to:
=
factoring:
=
=
QED.
(Assuming (*) is true for n=k leads to (*) being true for n=k+1)
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