A general binomial

Probability Level 4

( 1 + x ) n = ∑ r = 0 n ( n r ) x r {(1+x)}^n = \displaystyle \sum_{r=0}^n \dbinom nr x^r

Given the binomial expansion above, evaluate the value of

2 [ ( n 0 ) + ( n 3 ) + ( n 6 ) + ⋯ + ( n n ) ] + ( 1 + ω ) [ ( n 1 ) + ( n 4 ) + ( n 7 ) + ⋯ + ( n n − 2 ) ] + ( 1 + ω 2 ) [ ( n 2 ) + ( n 5 ) + ( n 8 ) + ⋯ + ( n n − 1 ) ] 2 \left[ \dbinom n0 + \dbinom n3 + \dbinom n6 + \cdots + \dbinom nn \right] \\ + (1 + \omega) \left[ \dbinom n1 + \dbinom n4 + \dbinom n7 + \cdots + \dbinom{n}{n-2} \right] \\ + (1 + \omega^2) \left[ \dbinom n2 + \dbinom n5 + \dbinom n8 + \cdots + \dbinom{n}{n-1} \right]

where

  • ω \omega denotes the non-real cube root of unity.
  • n n in an odd integral multiple of 3.
  • ( M N ) = M ! N ! ( M − N ) ! \dbinom MN = \dfrac {M!}{N! (M-N)!} denotes the binomial coefficient .
2 n 2^n 2 n + 1 2^n +1 0 0 2 n − 1 2^n -1

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Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Tapas Mazumdar
Jan 29, 2017

When x = ω x=\omega

( 1 + ω ) n = ( n 0 ) + ( n 1 ) ω + ( n 2 ) ω 2 + . . . + ( n n ) ω n {(1+\omega)}^n = \dbinom n0 + \dbinom n1 \omega + \dbinom n2 \omega^2 + ... + \dbinom nn \omega^n

When x = 1 x=1

( 1 + 1 ) n = ( n 0 ) + ( n 1 ) + ( n 2 ) + . . . + ( n n ) {(1+1)}^n = \dbinom n0 + \dbinom n1 + \dbinom n2 + ... + \dbinom nn

Adding the above two equation gives

( 1 + ω ) n + 2 n = 2 ( n 0 ) + ( 1 + ω ) ( n 1 ) + ( 1 + ω 2 ) ( n 2 ) + 2 ( n 3 ) + ( 1 + ω ) ( n 4 ) + ( 1 + ω 2 ) ( n 5 ) + 2 ( n 6 ) + . . . + ( 1 + ω ) ( n n − 2 ) + ( 1 + ω 2 ) ( n n − 1 ) + 2 ( n n ) As 1 + ω 3 k = 2 \begin{aligned} {(1+\omega)}^n + 2^n &=& 2 \dbinom n0 + (1+\omega) \dbinom n1 + (1+\omega^2) \dbinom n2 + 2 \dbinom n3 + \\ \\ & & (1+\omega) \dbinom n4 + (1+\omega^2) \dbinom n5 + 2 \dbinom n6 + ... + \\ \\ & & (1+\omega) \dbinom{n}{n-2} + (1+\omega^2) \dbinom{n}{n-1} + 2 \dbinom nn \qquad \qquad \small \color{#3D99F6}{\text{As } 1+\omega^{3k} = 2} \end{aligned} \\

∴ 2 n + ( − ω 2 ) n = 2 [ ( n 0 ) + ( n 3 ) + ( n 6 ) + . . . + ( n n ) ] + ( 1 + ω ) [ ( n 1 ) + ( n 4 ) + ( n 7 ) + . . . + ( n n − 2 ) ] + ( 1 + ω 2 ) [ ( n 2 ) + ( n 5 ) + ( n 8 ) + . . . + ( n n − 1 ) ] ⟹ 2 n − 1 = 2 [ ( n 0 ) + ( n 3 ) + ( n 6 ) + . . . + ( n n ) ] + ( 1 + ω ) [ ( n 1 ) + ( n 4 ) + ( n 7 ) + . . . + ( n n − 2 ) ] + ( 1 + ω 2 ) [ ( n 2 ) + ( n 5 ) + ( n 8 ) + . . . + ( n n − 1 ) ] As n is an odd integral multiple of 3 so ( − ω 2 ) n = 1 \\ \begin{aligned} \therefore \qquad & 2^n + {(-\omega^2)}^n &= 2 \left[ \dbinom n0 + \dbinom n3 + \dbinom n6 + ... + \dbinom nn \right] \\ \\ & & + (1 + \omega) \left[ \dbinom n1 + \dbinom n4 + \dbinom n7 + ... + \dbinom{n}{n-2} \right] \\ \\ & & + (1 + \omega^2) \left[ \dbinom n2 + \dbinom n5 + \dbinom n8 + ... + \dbinom{n}{n-1} \right] \\ \\ \\ \\ \implies \qquad & \boxed{2^n - 1} &= 2 \left[ \dbinom n0 + \dbinom n3 + \dbinom n6 + ... + \dbinom nn \right] \\ \\ & & + (1 + \omega) \left[ \dbinom n1 + \dbinom n4 + \dbinom n7 + ... + \dbinom{n}{n-2} \right] \\ \\ & & + (1 + \omega^2) \left[ \dbinom n2 + \dbinom n5 + \dbinom n8 + ... + \dbinom{n}{n-1} \right] \qquad \qquad \small \color{#3D99F6}{\text{As } n \text{ is an odd integral multiple of } 3 \text{ so } {(-\omega^2)}^n = 1} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Doesn't it also matter whether n is even or odd? Or am I missing something? You specify that n is a multiple of 3, but not that it's odd. Looks to me as though you should get 2^n + 1 if n were even.

Steve McMath - 4 years, 4 months ago

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Oh yes, I see now. Thanks for pointing my mistake. I've made necessary corrections.

Tapas Mazumdar - 4 years, 4 months ago

This solution is wrong. 1 + ω = − ω 2 1 + \omega = -\omega^2 , so the correct simplification is ( 1 + ω ) n + 2 n = ( − ω 2 ) n + 2 n (1 + \omega)^n + 2^n = (-\omega^2)^n + 2^n . Then, ω n = 1 \omega^n = 1 , so the answer is 2 n + ( − 1 ) n 2^n + (-1)^n .

EDIT: The problem originally just said that n n is a multiple of 3. With this change, we can now simplify ( − 1 ) n = − 1 (-1)^n = -1 because n n is odd, thus giving the given answer.

Ivan Koswara - 4 years, 4 months ago

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Oh yes, I see now. Thanks for pointing my mistake. I've made necessary corrections.

Tapas Mazumdar - 4 years, 4 months ago

Classic JEE solution :P, put n=3

A Former Brilliant Member - 4 years ago

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