$f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\},$ such
that for all $n=1,2,3,\ldots$ the sum $f(1)+f(2)+\cdots+f(n)$ is equal to
a perfect cube that is less than or equal to $n^3.$
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What values can
$f(1)$take? -
What value must
$f(1) + f(2)$take? -
Find an expression for
$\sum_{i=1}^{n}{f(i)}$in terms of$n.$ -
Prove your hypothesis using induction.
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What is the value of
$f(n)$if$\sum_{i=1}^{n}{f(i)} = n^3$and$\sum_{i=1}^{n-1}{f(i)} = (n-1)^3?$
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By trying small values of
$n,$we find that$f(1) \leq 1^3,$which means that necessarily$f(1)=1.$The next case gives us$f(1)+f(2) = 8.$This can give us an intuition that the function is unique:$f(n) = n^3 - (n-1)^3.$To prove this formally we use induction.Let
$\Phi(n)$be the induction hypothesis “If$S(i)=\sum_{j=1}^i{f(j)}$is a perfect cube less than or equal to$i^3$for all$i \leq n$then$S(i) = i^3.$”Base case:
$f(1) = 1$as it is the only perfect cube less than or equal to$1^3,$so$\Phi(1)$is true.Inductive case: Assuming
$\Phi(k)$is true, we can prove$\Phi(k+1)$as follows: Suppose$S(i)$is a perfect cube less than or equal to$i^3$for all$i \leq k+1.$Using the induction hypothesis$\Phi(k)$,$S(i) = i^3$for all$i \leq k.$Since$S(k) = k^3$and$S(k+1) \leq (k+1)^3,$$S(k+1)$must equal$(k+1)^3$as there are no cubes between$k^3$and$(k+1)^3.$Therefore$\Phi(k+1)$is true.By induction,
$S(n) = n^3$for all$n,$and so$f(n) = \sum_{j=1}^{n}{f(j)} - \sum_{j=1}^{n-1}{f(j)}$$= n^3 - (n-1)^3.$