Binomial Sums?

Algebra Level pending

Evaluate ∑ k = 0 n ( n k ) a k ( 1 − a ) n − k \large \sum_{k=0}^{n} \binom{n}{k}a^{k}(1-a)^{n-k}

Notation: ( n m ) = n ! m ! ( n − m ) ! \dbinom nm = \dfrac {n!}{m!(n-m)!} denotes the binomial coefficient .


The answer is 1.

James Watson by James Watson , 16, United Kingdom

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1 solution

James Watson
Jul 26, 2020

∑ k = 0 n ( n k ) a k ( 1 − a ) n − k ≡ ( a + ( 1 − a ) ) n \sum_{k=0}^{n} \binom{n}{k}a^{k}(1-a)^{n-k} \equiv \left( a + (1-a)\right)^n

( a + ( 1 − a ) ) n = 1 n = 1 \left( a + (1-a)\right)^n = 1^n = \boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The brackets ( ) for a k a^k and 1 n 1^n are unnecessary.

Chew-Seong Cheong - 10 months, 2 weeks ago

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ah yes, thanks for letting me know!

James Watson - 10 months, 2 weeks ago

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