To prove: P (u = 0) = p0
Solution:
= p_{0} B_{n,0} (u) + p_{1} B_{n, 1} (u) +...... + p_{n} B_{n, n}(u)...............(1)
B_{n,i} (u) = ^{n}c_{i} u^{i} (1 - u)^{n-i}
B_{n,0 }(u) = ^{n}c_{0} u^{0} (1 -u)^{n -0} = n!/(0!(n-0)!).1.(1 - u)^{n} = (1 - u)^{n}
B_{n,1} (u) = ^{n}c_{1} u^{1} (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 u^{i} (i = 0 to n) by using these terms along with u = 0 in (1) we find:
P (u = 0) = p_{0} (1 - 0)^{n} + p_{1}. 0. (1 - 0)^{n - 1} . n + 0 + 0 + ......+ 0
P (u = 0) = p_{0} Proved