To prove: P (u = 0) = p0

Solution:

= p₀ B_n,0 (u) + p₁ B_{n, 1} (u) +...... + p_n B_{n, n}(u)...............(1)

 B_n,i (u) = ⁿc_i uⁱ (1 - u)^n-i

B_n,0 (u) = ⁿc₀ u⁰ (1 -u)^{n -0} = n!/(0!(n-0)!).1.(1 - u)ⁿ = (1 - u)ⁿ

B_n,1 (u) = ⁿc₁ u¹ (1 -u)^{n -1} = n!/(1!(n-1)!).u.(1 - u)^n-1

We observe that all terms expect B_n,0  (u) have various of uⁱ  (i = 0 to n) by using these terms along with u = 0 in (1) we find:

P (u = 0) = p₀ (1 - 0)ⁿ + p₁. 0. (1 - 0)^{n - 1} . n + 0 + 0 + ......+ 0

P (u = 0) = p₀ Proved