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Question: Consider the binomial expansion of $(1+x)^n = ^nC_0 + ^nC_1x + ^nC_2x^2 + ... + ^nC_nx^n$. The value...

Consider the binomial expansion of (1+x)n=nC0+nC1x+nC2x2+...+nCnxn(1+x)^n = ^nC_0 + ^nC_1x + ^nC_2x^2 + ... + ^nC_nx^n. The value of 2⋅33C0−33C1−33C2+2⋅33C3−33C4−33C5+2⋅33C6−...+2⋅33C15−33C162 \cdot {^{33}C_0} - {^{33}C_1} - {^{33}C_2} + 2 \cdot {^{33}C_3} - {^{33}C_4} - {^{33}C_5} + 2 \cdot {^{33}C_6} - ... + 2 \cdot {^{33}C_{15}} - {^{33}C_{16}} is

A

-1

B

0

C

1

D

2162^{16}

Answer

-1

Explanation

Solution

Let the given sum be SS.

The general term in the sum follows a pattern of coefficients: 2,−1,−12, -1, -1 for CrC_r where r(mod3)r \pmod 3 is 0,1,20, 1, 2 respectively.

The sum can be written as:

S=(2⋅33C0−33C1−33C2)+(2⋅33C3−33C4−33C5)+⋯+(2⋅33C12−33C13−33C14)+2⋅33C15−33C16S = (2 \cdot {^{33}C_0} - {^{33}C_1} - {^{33}C_2}) + (2 \cdot {^{33}C_3} - {^{33}C_4} - {^{33}C_5}) + \dots + (2 \cdot {^{33}C_{12}} - {^{33}C_{13}} - {^{33}C_{14}}) + 2 \cdot {^{33}C_{15}} - {^{33}C_{16}}

This sum can be written as:

S=∑k=04(2⋅33C3k−33C3k+1−33C3k+2)+2⋅33C15−33C16S = \sum_{k=0}^{4} (2 \cdot {^{33}C_{3k}} - {^{33}C_{3k+1}} - {^{33}C_{3k+2}}) + 2 \cdot {^{33}C_{15}} - {^{33}C_{16}}

However, the term 2⋅33C152 \cdot {^{33}C_{15}} is already part of the sum 2⋅33C3k2 \cdot {^{33}C_{3k}} for k=5k=5.

So, let's rewrite the sum by carefully observing the last term.

The last term is 33C16{^{33}C_{16}}.

The pattern 2,−1,−12, -1, -1 applies to terms C3k,C3k+1,C3k+2C_{3k}, C_{3k+1}, C_{3k+2}.

The sum goes up to C16C_{16}.

16≡1(mod3)16 \equiv 1 \pmod 3.

So, the term C16C_{16} has coefficient −1-1.

The term C15C_{15} (since 15≡0(mod3)15 \equiv 0 \pmod 3) has coefficient 22.

The term C14C_{14} (since 14≡2(mod3)14 \equiv 2 \pmod 3) has coefficient −1-1.

The sum is:

S=(2C0−C1−C2)+(2C3−C4−C5)+(2C6−C7−C8)+(2C9−C10−C11)+(2C12−C13−C14)+2C15−C16S = (2C_0 - C_1 - C_2) + (2C_3 - C_4 - C_5) + (2C_6 - C_7 - C_8) + (2C_9 - C_{10} - C_{11}) + (2C_{12} - C_{13} - C_{14}) + 2C_{15} - C_{16}.

Let's define a function f(r)f(r) such that f(r)=2f(r) = 2 if r≡0(mod3)r \equiv 0 \pmod 3, and f(r)=−1f(r) = -1 if r≡1(mod3)r \equiv 1 \pmod 3 or r≡2(mod3)r \equiv 2 \pmod 3.

The given sum is ∑r=016f(r)33Cr\sum_{r=0}^{16} f(r) {^{33}C_r}.

Consider the properties of cube roots of unity. Let ω=ei2π/3\omega = e^{i2\pi/3}. We know 1+ω+ω2=01+\omega+\omega^2=0.

Consider the sum T=∑r=0nnCrxrT = \sum_{r=0}^{n} {^{n}C_r} x^r.

We are interested in a sum of the form ∑r=0nkrnCr\sum_{r=0}^{n} k_r {^{n}C_r}.

Let's find a,b,ca, b, c such that kr=a+bωr+cω2rk_r = a + b\omega^r + c\omega^{2r}.

If r≡0(mod3)r \equiv 0 \pmod 3, kr=a+b+c=2k_r = a+b+c = 2.

If r≡1(mod3)r \equiv 1 \pmod 3, kr=a+bω+cω2=−1k_r = a+b\omega+c\omega^2 = -1.

If r≡2(mod3)r \equiv 2 \pmod 3, kr=a+bω2+cω4=a+bω2+cω=−1k_r = a+b\omega^2+c\omega^4 = a+b\omega^2+c\omega = -1.

Adding these three equations:

(a+b+c)+(a+bω+cω2)+(a+bω2+cω)=2−1−1=0(a+b+c) + (a+b\omega+c\omega^2) + (a+b\omega^2+c\omega) = 2 - 1 - 1 = 0

3a+b(1+ω+ω2)+c(1+ω+ω2)=03a + b(1+\omega+\omega^2) + c(1+\omega+\omega^2) = 0

3a+b(0)+c(0)=0  ⟹  3a=0  ⟹  a=03a + b(0) + c(0) = 0 \implies 3a = 0 \implies a = 0.

Substitute a=0a=0 into the equations:

b+c=2b+c = 2

bω+cω2=−1b\omega+c\omega^2 = -1

bω2+cω=−1b\omega^2+c\omega = -1

From b+c=2b+c=2, we have c=2−bc=2-b. Substitute into the second equation:

bω+(2−b)ω2=−1b\omega + (2-b)\omega^2 = -1

bω+2ω2−bω2=−1b\omega + 2\omega^2 - b\omega^2 = -1

b(ω−ω2)=−1−2ω2b(\omega-\omega^2) = -1 - 2\omega^2

Since ω2=−1−ω\omega^2 = -1-\omega:

b(ω−ω2)=−1−2(−1−ω)=−1+2+2ω=1+2ωb(\omega-\omega^2) = -1 - 2(-1-\omega) = -1+2+2\omega = 1+2\omega.

We know ω=−1/2+i3/2\omega = -1/2 + i\sqrt{3}/2 and ω2=−1/2−i3/2\omega^2 = -1/2 - i\sqrt{3}/2.

So ω−ω2=i3\omega-\omega^2 = i\sqrt{3}.

And 1+2ω=1+2(−1/2+i3/2)=1−1+i3=i31+2\omega = 1+2(-1/2+i\sqrt{3}/2) = 1-1+i\sqrt{3} = i\sqrt{3}.

Thus, b(i3)=i3  ⟹  b=1b(i\sqrt{3}) = i\sqrt{3} \implies b=1.

Since c=2−bc=2-b, c=2−1=1c=2-1=1.

So, the coefficients krk_r are given by kr=ωr+ω2rk_r = \omega^r + \omega^{2r}.

The sum ∑r=0nkrnCr=∑r=0n(ωr+ω2r)nCr\sum_{r=0}^{n} k_r {^{n}C_r} = \sum_{r=0}^{n} (\omega^r + \omega^{2r}) {^{n}C_r}

=∑r=0nnCrωr+∑r=0nnCrω2r= \sum_{r=0}^{n} {^{n}C_r} \omega^r + \sum_{r=0}^{n} {^{n}C_r} \omega^{2r}

=(1+ω)n+(1+ω2)n= (1+\omega)^n + (1+\omega^2)^n.

For n=33n=33:

(1+ω)33+(1+ω2)33(1+\omega)^{33} + (1+\omega^2)^{33}

Since 1+ω=−ω21+\omega = -\omega^2 and 1+ω2=−ω1+\omega^2 = -\omega:

=(−ω2)33+(−ω)33= (-\omega^2)^{33} + (-\omega)^{33}

=−(ω2)33−(ω)33= -(\omega^2)^{33} - (\omega)^{33} (since 33 is odd)

=−(ω66)−(ω33)= -(\omega^{66}) - (\omega^{33})

Since ω3=1\omega^3=1:

=−(ω3)22−(ω3)11= -(\omega^3)^{22} - (\omega^3)^{11}

=−(1)22−(1)11=−1−1=−2= -(1)^{22} - (1)^{11} = -1 - 1 = -2.

This result, −2-2, is the value of the sum ∑r=033f(r)33Cr\sum_{r=0}^{33} f(r) {^{33}C_r}.

Let this full sum be SfullS_{full}.

Sfull=f(0)C0+f(1)C1+⋯+f(16)C16+⋯+f(33)C33S_{full} = f(0)C_0 + f(1)C_1 + \dots + f(16)C_{16} + \dots + f(33)C_{33}.

Sfull=(2C0−C1−C2)+(2C3−C4−C5)+⋯+(2C30−C31−C32)+2C33S_{full} = (2C_0 - C_1 - C_2) + (2C_3 - C_4 - C_5) + \dots + (2C_{30} - C_{31} - C_{32}) + 2C_{33}.

The last term is 2C332C_{33} because 33≡0(mod3)33 \equiv 0 \pmod 3.

The given sum is S=2C0−C1−C2+⋯+2C15−C16S = 2C_0 - C_1 - C_2 + \dots + 2C_{15} - C_{16}.

Let's express SfullS_{full} in terms of SS.

Sfull=∑r=033f(r)33Cr=∑r=016f(r)33Cr+∑r=1733f(r)33CrS_{full} = \sum_{r=0}^{33} f(r) {^{33}C_r} = \sum_{r=0}^{16} f(r) {^{33}C_r} + \sum_{r=17}^{33} f(r) {^{33}C_r}.

So Sfull=S+∑r=1733f(r)33CrS_{full} = S + \sum_{r=17}^{33} f(r) {^{33}C_r}.

We use the property nCr=nCn−r{^nC_r} = {^nC_{n-r}}. Here n=33n=33.

S=∑r=016f(r)33CrS = \sum_{r=0}^{16} f(r) {^{33}C_r}.

Consider Srev=∑r=1733f(r)33CrS_{rev} = \sum_{r=17}^{33} f(r) {^{33}C_r}.

Let k=33−rk=33-r. When r=17,k=16r=17, k=16. When r=33,k=0r=33, k=0.

Srev=∑k=016f(33−k)33C33−k=∑k=016f(33−k)33CkS_{rev} = \sum_{k=0}^{16} f(33-k) {^{33}C_{33-k}} = \sum_{k=0}^{16} f(33-k) {^{33}C_k}.

Since 33≡0(mod3)33 \equiv 0 \pmod 3, we have 33−k≡−k(mod3)33-k \equiv -k \pmod 3.

If k≡0(mod3)k \equiv 0 \pmod 3, then 33−k≡0(mod3)33-k \equiv 0 \pmod 3. So f(33−k)=2=f(k)f(33-k) = 2 = f(k).

If k≡1(mod3)k \equiv 1 \pmod 3, then 33−k≡−1≡2(mod3)33-k \equiv -1 \equiv 2 \pmod 3. So f(33−k)=−1=f(k)f(33-k) = -1 = f(k).

If k≡2(mod3)k \equiv 2 \pmod 3, then 33−k≡−2≡1(mod3)33-k \equiv -2 \equiv 1 \pmod 3. So f(33−k)=−1=f(k)f(33-k) = -1 = f(k).

Thus, f(33−k)=f(k)f(33-k) = f(k) for all kk.

So, Srev=∑k=016f(k)33Ck=SS_{rev} = \sum_{k=0}^{16} f(k) {^{33}C_k} = S.

Therefore, Sfull=S+Srev=S+S=2SS_{full} = S + S_{rev} = S+S = 2S.

We found Sfull=−2S_{full} = -2.

So, 2S=−22S = -2.

S=−1S = -1.

The final answer is -1\boxed{\text{-1}}.