Using Binomial Theorem, give the closed form expression for: ∑k=0n\textstyle\sum_{k=0}^n∑k=0n(nk){n \choose k}(kn)3n · 2k
Using Binomial Theorem, let us give the closed form expression for
∑k=0n(nk)3n⋅2k.\sum\limits_{k=0}^n {n \choose k}3^n · 2^k.k=0∑n(kn)3n⋅2k.
It follows that
∑k=0n(nk)3n⋅2k=3n∑k=0n(nk)2k1n−k=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.k=0∑n(kn)3n⋅2k=3nk=0∑n(kn)2k1n−k=3n(2+1)n=3n3n=32n=9n.
We conclude that the closed form expression is 9n.9^n.9n.
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