Problem on Mathematical Induction

Question

How does one solve $f:=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105},\ \forall k \in\mathbb{N}, \ \exists f\in\mathbb{N}$?

Answer 1

Inductive step: We have $$P (k+1) =\frac {(1+k)^7}{7} +\frac {(1+k)^5}{5} +\frac {2 (k+1)^3}{3} -\frac {1+k}{105} $$ $$= \frac {k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1}{7} +\frac {k^5+5k^4+10k^3+5k^4+1}{5} +\frac {2k^3+6k^2+6k+2}{3} -\frac {1+k}{105} $$ $$=[\frac {k^7}{7} +\frac {k^5}{5} +\frac {2k^3}{3} -\frac {k}{105}] +[k^6+3k^5+5k^4+5k^3+3k^2+k]+[k^4+2k^3+k^4]+[2k^2+2k]+[\frac {1}{7}+\frac {1}{5}+\frac {2}{3}-\frac{1}{105}] $$ $$=\text{Integer} + \text{Integer} + \text{Integer} + \text{Integer} +1 $$ $$=\text{Integer} $$ We are done!! Hope it helps.

Answer 2

Here is a method not using induction.

Over a common denominator we have $$\frac{15k^7+21k^5+70k^3-k}{105}$$

Modulo $7$ the numerator reduces to $k^7-k$, modulo $5$ we get $k^5-k$ and modulo $3$ it is $k^3-k$. It is a simple application of Fermat's little theorem to show that the numerator is therefore divisible by $7\times 5 \times 3=105$ and the expression is an integer.

Answer 3

The question states:

Prove by induction that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is integer.

Let's rephrase the question:

Prove by induction that $15k^7+21k^5+70k^3-k$ is divisible by $105$.

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First, show that this is true for $k=1$:

$15\cdot1^7+21\cdot1^5+70\cdot1^3-1=105$

Second, assume that this is true for $k$:

$15k^7+21k^5+70k^3-k=105n$

Third, prove that this is true for $k+1$:

$15(k+1)^7+21(k+1)^5+70(k+1)^3-(k+1)=$

$\color\red{15k^7+21k^5+70k^3-k}+105(k^6+3k^5+6k^4+7k^3+7k^2+4k+1)=$

$\color\red{105n}+105(k^6+3k^5+6k^4+7k^3+7k^2+4k+1)=$

$105(n+k^6+3k^5+6k^4+7k^3+7k^2+4k+1)$

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Please note that the assumption is used only in the part marked red.