Let p be a prime and m a positive integer. By mathematical induction on m, or otherwise, prove that whenever r is an integer such that p does not divide r, p divides $^{m p} C_{r}$

Open in App

Solution

$^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1}$
$∴^{m} p C_{r} = {\frac{m p}{r}}^{m p - 1} C_{r - 1} = [\frac{m .^{m p - 1} C_{r - 1}}{r}]$
L.H.S. is an integer $\Rightarrow$ R.H.S. is an integer and p is prime such that p does not divide r.
$∴ \frac{^{m} p C_{r}}{p}$ = integer which is turn means that p divides $^{m} p C_{r} .$
Proof by Introduction :
For m = 1, $^{m} p C_{r} =^{p} C_{r}$
$= \frac{p (p - 1) (p - 2) . . . . r f a c t o r s}{r (r - 1) (r - 2) . . . . r f a c t o r s}$
Above is clearly divisible by p.(r < p)
Assume $^{m p} C_{r}$ is divisible by p.
We shall prove that $^{(m + 1) p} C_{r}$ is also divisible by p to complete the proof.
Now $(1 + x)^{(m + 1) p} = (1 + x)^{m p} (1 + x)^{p}$
Expanding both sides and equating the coefficient of $x^{r}$ on both sides, we get
$^{(m + 1) p} C_{r} = 1.^{p} C_{r} +^{m p} C_{1}^{p} C r - 1 +^{m p} C_{2}^{p} C_{r - 2} + . . . . .$ R.H.S. is clearly divisible by p by (1) and (2) and hence L.H.S. i.e., $^{(m + 1) p} C_{r}$ is also divisible by p.

Suggest Corrections

Similar questions

\sqrt{n}

Let p be a prime number. If p divides $a^{2}$ then _________, where a is a positive integer.

a^{2}

Let p be a prime number. If p divides $a^{2}$ then p divides a, where a is a positive integer. Which theorem is this statement previously based on?

Join BYJU'S Learning Program