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[homeschool girl]

For some positive integer [MATH]n,[/MATH] the expansion of [MATH](1 + x)^n[/MATH] has three consecutive coefficients [MATH]a,[/MATH] [MATH]b,[/MATH] [MATH]c[/MATH] that satisfy [MATH]a:b:c = 1:7:35.[/MATH] What must [MATH]n[/MATH] be?

I think i need to solve it by using the binomial theorem like this:

[MATH]7\cdot\binom{n}{r}\cdot1^{n-r}\cdot x^r=\binom{n}{r+1}\cdot1^{n-(r+1)}\cdot x^{r+1}[/MATH]
[MATH]35\cdot\binom{n}{r}\cdot1^{n-r}\cdot x^r=\binom{n}{r+2}\cdot1^{n-(r+2)}\cdot x^{r+2}[/MATH]
but I'm not sure. can someone please help?

 

[Deleted member 4993]

For some positive integer [MATH]n,[/MATH] the expansion of [MATH](1 + x)^n[/MATH] has three consecutive coefficients [MATH]a,[/MATH] [MATH]b,[/MATH] [MATH]c[/MATH] that satisfy [MATH]a:b:c = 1:7:35.[/MATH] What must [MATH]n[/MATH] be?

I think i need to solve it by using the binomial theorem like this:

[MATH]7\cdot\binom{n}{r}\cdot1^{n-r}\cdot x^r=\binom{n}{r+1}\cdot1^{n-(r+1)}\cdot x^{r+1}[/MATH]
[MATH]35\cdot\binom{n}{r}\cdot1^{n-r}\cdot x^r=\binom{n}{r+2}\cdot1^{n-(r+2)}\cdot x^{r+2}[/MATH]
but I'm not sure. can someone please help?

What is the COEFFICIENT of the kth term of (1+x)ⁿ?

What is the COEFFICIENT of the (k+1) th term of (1+x)ⁿ?

What is the COEFFICIENT of the (k- 1) th term of (1+x)ⁿ?

 

[homeschool girl]

ohhhhh it would just be [MATH]7\cdot\binom{n}{k}=\binom{n}{k+1},[/MATH] and [MATH]35\cdot\binom{n}{k}=\binom{n}{k+2}[/MATH] because we're only looking for the coefficients

 

[Deleted member 4993]

ohhhhh it would just be [MATH]\binom{n}{k},[/MATH] [MATH]\binom{n}{k+1},[/MATH] and [MATH]\binom{n}{k-1}[/MATH] because we're only looking for the coefficients

Yes - now the real hard part. Now choose a pair and take the ratio and put those equal to the given number ratios.

You should get equations that will help you solve for 'n'.

 

[homeschool girl]

I got a system of equations but I don't know how to solve it

 

[Deleted member 4993]

I got a system of equations but I don't know how to solve it

Please share your work so that we know how/where to help you.

 

[homeschool girl]

[MATH]\frac{7\cdot n!}{k!\cdot(n-k)!}=\frac{n!}{(k+1)!\cdot(n-(k+1))!}[/MATH]
[MATH]\frac{35\cdot n!}{k!\cdot(n-k)!}=\frac{n!}{(k+2)!\cdot(n-(k+2))!}[/MATH]

 

[homeschool girl]

i do know that i'm supposed to use [MATH]\frac{\binom{n}{k}}{\binom{n}{k - 1}}=\frac{n-k+1}{k}[/MATH] to help me but i dont know where to use it

 

[Deleted member 4993]

[MATH]\frac{7\cdot n!}{k!\cdot(n-k)!}=\frac{n!}{(k+1)!\cdot(n-(k+1))!}[/MATH]
[MATH]\frac{35\cdot n!}{k!\cdot(n-k)!}=\frac{n!}{(k+2)!\cdot(n-(k+2))!}[/MATH]

[MATH]7 =\frac{\frac{n!}{(k+1)!\cdot(n-(k+1))!}}{\frac{ n!}{k!\cdot(n-k)!}}[/MATH]
[MATH]7 =\frac{k!\cdot(n-k)!}{(k+1)!\cdot(n-(k+1))!}[/MATH]
[MATH]7 =\frac{(n-k)!}{(k+1)\cdot(n-(k+1))!}[/MATH]
[MATH]7 =\frac{(n-k)!}{(k+1)\cdot(n-(k+1)!}[/MATH]
[MATH]7 =\frac{(n-k)}{(k+1)}[/MATH]........ continue.....

 

[homeschool girl]

I don't quite understand how you got [MATH]7 =\frac{\frac{n!}{(k+1)!\cdot(n-(k+1))!}}{\frac{ n!}{k!\cdot(n-k)!}}.[/MATH]

 

[Deleted member 4993]

I don't quite understand how you got [MATH]7 =\frac{\frac{n!}{(k+1)!\cdot(n-(k+1))!}}{\frac{ n!}{k!\cdot(n-k)!}}.[/MATH]

Really! It is same as:

7 * A = B .... coming from your first equation...\(\displaystyle \ \to \ \ \ 7 = \ \frac{B}{A} \)

 

[homeschool girl]

Thank you @Subhotosh Khan it helped a ton, I appreciate it

 

[Deleted member 4993]

Thank you @Subhotosh Khan it helped a ton, I appreciate it

Did you solve the problem?

What was the value of 'k'?

 

[homeschool girl]

I got k=2 and n=23 (i also got -1 and -1, but when I checked it made a denominator equal to zero)

 

[Deleted member 4993]

I got k=2 and n=23 (i also got -1 and -1, but when I checked it made a denominator equal to zero)

Excellent!

Another way to discard, k=-1 and n=-1, would be to think the physical meaning of n and k in combination problems. It is 'k' choosing from 'n' available states - those cannot be negative.