In a population of diploid organisms If frequency of allele A = p and frequency of allele a = q

Expected genotype frequency under random mating are

AA = p² (for the AA homozygotes)

aa = q² (for the aa homozygotes)

Aa = 2pq (for the Aa heterozygotes)

(In absence of selection , mutation , genetic drift or other forces allelic frequency p and q are constant through generation) Therefore p² + 2pq +q² = 1

Detailed Answer :

The Hardy-Weinberg model enables us to compare a population's actual genetic structure over time with the genetic structure we would expect if the population were in Hardy-Weinberg equilibrium (i.e., not evolving). If genotype frequencies differ from those we would expect under equilibrium, we can assume that one or more of the model's assumptions are being violated, and attempt to determine which one(s).

Hardy and Weinberg assigned the letter p to the frequency of the dominant allele A and the letter q to the frequency of the recessive allele a.

Since the sum of all the alleles must equal 100%, then p + q = 1. They then reasoned that all the  random possible combinations of the members of a population would equal (p+q)² or p² + 2pq + q² .

The overall equation for the Hardy-Weinberg equilibrium is expressed in this way: p² + 2pq + q² = 1 [binomial expansion of (p + q)²]