Question from Trial (1 Viewer)

[lyounamu]

For k = 0, 1, 2, 3, 4, 5, ..., n, Pk (k term) is defined as:
Pk = nCk . a^k . (1-a)^(n-k) where a is real and n>0.

Prove that "Sigma notation" (k=0 on the bottom with n on top) Pk = 1


I just tried it and I cannot seem to get my mind around this. Any helpn will be greatly appreciated. Thanks in advance.

 

[duy.le]

ok my solution is kinda weird but it makes sense.

Pk=nCk.a^k.(1-a)^(n-k)............................(1)

so what i did was think about what that represented and realised that that was the Kth term in a binomial expansion. so hence

the sum of the terms Pk, ie P0+P1+P2+....+Pn would collapse back into a binomial of two terms to the power of n; ie. (a+b)^n

from the general term

Pk=nCk.a^n.b^(n-k) we just now substitute back, comparing to (1)

hence the sum of Pk from 0->n = (a+(1-a))^n = 1

-=QED=-

 

[lyounamu]

ok my solution is kinda weird but it makes sense.

Pk=nCk.a^k.(1-a)^(n-k)............................(1)

so what i did was think about what that represented and realised that that was the Kth term in a binomial expansion. so hence

the sum of the terms Pk, ie P0+P1+P2+....+Pn would collapse back into a binomial of two terms to the power of n; ie. (a+b)^n

from the general term

Pk=nCk.a^n.b^(n-k) we just now substitute back, comparing to (1)

hence the sum of Pk from 0->n = (a+(1-a))^n = 1

-=QED=-

Ah, ok. I didn't see that. Thanks again.