Binomial quotient limit

Algebra Level 5

lim n k = 0 n ( n k ) k = 0 n / m ( n m k ) \Large {\lim_{n \to \infty}} {\dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{n/m}_{k=0}\dbinom{n}{mk}}}

If n n is a multiple of some positive integer m m , find the limit above.

m m Cannot be determined 0 0 2 m 2m 1 1 \infty

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1 solution

Akeel Howell
Mar 5, 2017

k = 0 n ( n k ) = ( 1 + 1 ) n = 2 n \displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k} = (1+1)^n = 2^n} k = 0 n / m ( n m k ) = ( 1 + ϕ ) n + ( 1 + ϕ 2 ) n + + ( 1 + ϕ m 1 ) n + ( 1 + 1 ) n m \large\displaystyle{\sum^{n/m}_{k=0}\dbinom{n}{mk} = \dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+(1+1)^n}{m}} = ( 1 + ϕ ) n + ( 1 + ϕ 2 ) n + + ( 1 + ϕ m 1 ) n + 2 n m = \large\dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n}{m} k = 0 n ( n k ) k = 0 n / m ( n m k ) = m × 2 n ( 1 + ϕ ) n + ( 1 + ϕ 2 ) n + + ( 1 + ϕ m 1 ) n + 2 n \implies \dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{n/m}_{k=0}\dbinom{n}{mk}} = \large\dfrac{m \times 2^n}{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n} lim n k = 0 n ( n k ) k = 0 n / m ( n m k ) = lim n ( m × 2 n 2 n ) ( ( 1 + ϕ ) n + ( 1 + ϕ 2 ) n + + ( 1 + ϕ m 1 ) n + 2 n 2 n ) = m \therefore \lim_{n \to \infty}{\dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{n/m}_{k=0}\dbinom{n}{mk}} = \large\lim_{n \to \infty}{\dfrac{\left(\dfrac{m \times 2^n}{2^n}\right)}{\left(\dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n}{2^n}\right)} = \boxed{m}}}


Note: Here, ϕ \large\phi is an m t h m^{th} root of unity.

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