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(a) Simplify
$$\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}.$$
(b) For some positive integer $$n$$ the expansion of $$(1 + x)^n$$ has three consecutive coefficients  $$a, b, c$$  that satisfy $$a:b:c = 1:8:40.$$ What must $$n$$ be?

Apr 26, 2023

#1
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(a) $$\frac{\binom{n}{k}}{\binom{n}{k - 1}} = \frac{\frac{n!}{k! (n - k)!}}{\frac{n!}{(k - 1)! (n - k + 1)!}} = \frac{n!(k - 1)! (n - k + 1)!}{k! (n - k)! n!} = \frac{(k - 1)(n - k + 1)}{k}$$

(b)

Let the expansion of (1 + x)^n be:

(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n)x^n

We know that:

a:b:c = 1:8:40

Therefore:

C(n, 0) : C(n, 1) : C(n, 2) = 1 : 8 : 40

Using the formula for the binomial coefficients, we have:

C(n, 0) = 1 C(n, 1) = n

Therefore:

C(n, 2) = 28C(n, 1) - 27C(n, 0)

Using the values we have:

C(n, 2) = 28n - 27

We know that:

C(n, 0) + C(n, 1) + C(n, 2) = 1 + 8 + 40 = 49

Therefore:

1 + n + C(n, 2) = 49

Solving this equation, we get n = 9.

Apr 26, 2023
#2
+1

b)

Suppose the coefficients are (n C r−1), (n C r), (n C r+1).

Then: [n - r + 1] / r ==8 / 1
[n - r] / [r +1] ==40 / 8, solve for n, r
n ==17  and  r ==2

So, the coefficients are: 17 C 1 : 17 C 2 : 17 C 3==
17 : 136 : 680 ==1 : 8 : 40

Apr 26, 2023
#8
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This is nonsense!  Check your math!

Guest Apr 26, 2023
#3
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The expansion of (1+x)n is (0n​)+(1n​)x+(2n​)x2+(3n​)x3+⋯. The three consecutive coefficients that satisfy a:b:c=1:8:40 are (0n​), (1n​), and (2n​). Therefore, we have \begin{align*} \frac{\binom{n}{2}}{\binom{n}{1}}&=\frac{8}{\binom{n}{0}}\ \frac{\binom{n}{2}}{\binom{n}{0}}&=8\ \binom{n}{2}&=8\binom{n}{0}\ \frac{n!(n-2)!}{2!(n-2)!}&=8\cdot\frac{n!}{n!}\ \frac{n!}{2!}&=8\ 2^n&=2^3\ n&=\boxed{3} \end{align*}

So n = 3

Apr 26, 2023
#4
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Expand (1 + x)^3 = x^3 + 3 x^2 + 3 x + 1

With n=3, how do you get coefficients to be in the ratio: 1 : 8 : 40

Guest Apr 26, 2023
#5
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(b) We know that the expansion of (1 + x)^n is given by the binomial theorem:

(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n)x^n

We are given that there exist three consecutive coefficients a, b, c such that a:b:c = 1:8:40. Let's denote these coefficients as C(n, k), C(n, k + 1), and C(n, k + 2), where k is some integer. Then we have:

a:b = C(n, k):C(n, k + 1) = 1:8

b:c = C(n, k + 1):C(n, k + 2) = 8:40 = 1:5

Multiplying these two ratios, we get:

a:b:c = 1:8:40

Therefore, we have:

C(n, k) = a

C(n, k + 1) = 8a

C(n, k + 2) = 40a

Using the formula for the binomial coefficients, we can express these coefficients in terms of n and k:

C(n, k) = n! / (k!(n - k)!)

C(n, k + 1) = n! / ((k + 1)!(n - k - 1)!)

C(n, k + 2) = n! / ((k + 2)!(n - k - 2)!)

We can then use these equations to eliminate a and simplify the ratios:

a:b:c = 1:8:40 n! / (k!(n - k)!) : n! / ((k + 1)!(n - k - 1)!) : n! / ((k + 2)!(n - k - 2)!) = 1:8:40

Simplifying this equation, we get:

(k + 2)(k + 1) = 40(n - k)

Expanding the left side and simplifying, we get:

k^2 + 3k - 38n + 40 = 0

We can use the quadratic formula to solve for k:

k = (-3 ± sqrt(9 + 152n)) / 2

Since k is an integer, we need the discriminant to be a perfect square:

9 + 152n = m^2

Solving for n, we get:

n = (m^2 - 9) / 152

Since n is an integer, m must be odd. Trying odd values of m, we find that the smallest value that works is m = 13, which gives:

n = (13^2 - 9) / 152 = 16

Apr 26, 2023
#6
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Do your math again:  n = (13^2 - 9) / 152 = 16

How in the world you got 16 from that ?! Besides: Expand: ( 1 + x)^16 = x^16 + 16 x^15 + 120 x^14 + 560 x^13 + 1820 x^12 + 4368 x^11 + 8008 x^10 + 11440 x^9 + 12870 x^8 + 11440 x^7 + 8008 x^6 + 4368 x^5 + 1820 x^4 + 560 x^3 + 120 x^2 + 16 x + 1

How do you the ratio of: 1 : 8 : 40 from your solution?

Guest Apr 26, 2023
#7
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(b) To find n, we can use the fact that a:b:c = 1:8:40, which means that the ratio of consecutive binomial coefficients is 1:8:40. Let's say that the middle coefficient is C(n,k). Then we have:

C(n, k-1) / C(n,k) = 1/8 C(n, k) / C(n, k+1) = 8/40 = 1/5

Using the identity C(n,k) = n! / (k!(n-k)!), we can simplify the expressions above as follows:

(k/(n-k+1)) / [(n-k)/(k+1)] = 1/8 [(n-k)/(k+1)] / [(n-k+1)/(k+2)] = 1/5

Solving these equations simultaneously will give us the values of k and n. However, this may involve some trial and error. Here is one possible approach:

From the first equation, we have k = (n-7)/9.

Substituting k into the second equation and simplifying, we get:

(n-15)(n-14) = 0

Therefore, n = 15 or n = 14.

We can then use k = (n-7)/9 to find the corresponding value of k:

For n=15, we have k = (15-7)/9 = 8/9. This is not an integer, so it does not work.

For n=14, we have k = (14-7)/9 = 7/9. This is not an integer either, so it does not work.

Therefore, there is no positive integer n that satisfies the given condition.

Apr 26, 2023
#9
+214
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(a): The given expression is equal to $$\frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k-1)!(n-k+1)!}}$$ .

Cancel n! to get $$\frac{(k-1)!(n-k+1)}{k!(n-k)!}$$

Cancel $$(k-1)!(n-k+1)!$$ to get $$\frac{1}{k(n-k)}$$.

(b): I agree that the answer is 17.

May 2, 2023