Theorem: If P is Prime then all the coefficients in (a + b)^P except the first and last are divisible by P.
Proof;
Let suppose P is prime then by definition it is only divisible by 1 and itself.
 |
| If p is prime, all the coefficients in (a+b)^p except the first and last are divisible by p. |
By Binomial Theorem
(a + b)^P = PC0* a^p * b^0 + PC1* a^p-1 * b^1 + PC2 * a^p-2 * b^2 + ...+ PCP-1 * a^1 * b^p-1 + PCP * a^0 * b^p
Where PCr denotes the binomial coefficients and is given by
PCr = P! / r!*(P - r)!
Since PCr is also the coefficients of rth term in the binomial expansion of (a + b)^P .
We have to show that for all values of r except when r=0 or r=P the Coefficients PCr of rth term in expansion is divisible by P.
Since P is only divisible by 1 and itself. So for 0 < k < P there is no P as common factor neither in r! nor in (P – r)! .
This implies that PCr = P! / r!*(P - r)! has no P as common factor in denominator which can cancel out factor P of P! in numerator.
Hence proved that if p is prime then all the coefficients in the expansion of (a+b)^p except the first and last are divisible by P.