Answer to Question #261050 in Discrete Mathematics for Alina

Question #261050

Using Binomial Theorem, give the closed form expression for: k=0n\textstyle\sum_{k=0}^n(nk){n \choose k}3n · 2k

1
Expert's answer
2021-11-04T19:58:52-0400

Using Binomial Theorem, let us give the closed form expression for

k=0n(nk)3n2k.\sum\limits_{k=0}^n {n \choose k}3^n · 2^k.


It follows that

k=0n(nk)3n2k=3nk=0n(nk)2k1nk=3n(2+1)n=3n3n=32n=9n.\sum\limits_{k=0}^n {n \choose k}3^n · 2^k =3^n\sum\limits_{k=0}^n {n \choose k}2^k1^{n-k}\\=3^n(2+1)^n=3^n3^n=3^{2n}=9^n.


We conclude that the closed form expression is 9n.9^n.


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