Binoomial theorem question (1 Viewer)

[vds700]

Prove that r .ⁿC_r = n . ^n-1C_r-1. Im fine with this, just using the definition for nCr

If P_r = ⁿC_r x^r (1 - x)^{n - r}, find the value of P₁ + P₂ + ...+ P_n.

Thanks for any help

 

[3unitz]

Prove that r .ⁿC_r = n . ^n-1C_r-1. Im fine with this, just using the definition for nCr

If P_r = ⁿC_r x^r (1 - x)^{n - r}, find the value of P₁ + P₂ + ...+ P_n.

Thanks for any help

P1 = ⁿC₁ x (1-x)^n-1
P2 = ⁿC₂ x² (1-x)^n-2
P3 = ⁿC₃ x³ (1-x)^n-3
...

notice the similarity to the expansion of:

(a + b)^n = a^n + ⁿC₁ a^n-1 b + ⁿC₂ a^n-2b² etc

our a corresponds with (1-x) and our b corresponds with x except that we're missing the first term P0 (which would be (1-x)^n)

P0 + P1 + P2 + P3 + ... + Pn = [x + (1-x)]^n
P0 + P1 + P2 + P3 + ... + Pn = 1
P1 + P2 + P3 + ... + Pn = 1 - (1-x)^n

 

[vds700]

P1 = ⁿC₁ x (1-x)^n-1
P2 = ⁿC₂ x² (1-x)^n-2
P3 = ⁿC₃ x³ (1-x)^n-3
...

notice the similarity to the expansion of:

(a + b)^n = a^n + ⁿC₁ a^n-1 b + ⁿC₂ a^n-2b² etc

our a corresponds with (1-x) and our b corresponds with x except that we're missing the first term P0 (which would be (1-x)^n)

P0 + P1 + P2 + P3 + ... + Pn = [x + (1-x)]^n
P0 + P1 + P2 + P3 + ... + Pn = 1
P1 + P2 + P3 + ... + Pn = 1 - (1-x)^n

thanks for your help. I wish id seen that, it seems so obvious now, i was trying to use the formula for the sum of a GP lol.