# Thread: How To Find The Constant And N Of Expanding (Pascal's Triangle)

1. ## How To Find The Constant And N Of Expanding (Pascal's Triangle)

I know the answer to these questions, but want to know how.

#1 The 10th term of the expansion of (x-(1/2))^n is -(1001/256)x^5

Find N.

The answer is 14. How do we get this? (I got it through trial and error, but I'm sure there is an easier way...)

#2 Which term in the expansion of ((1/(2x^2)) - x^3)^10 is a constant?

The answer is the 5th term.

I used trial and error for this one too. Though I"m sure there is an easier way too.

2. Originally Posted by AlphaRock
I know the answer to these questions, but want to know how.

#1 The 10th term of the expansion of (x-(1/2))^n is -(1001/256)x^5

Find N.

The answer is 14. How do we get this? (I got it through trial and error, but I'm sure there is an easier way...)

#2 Which term in the expansion of ((1/(2x^2)) - x^3)^10 is a constant?

The answer is the 5th term.

I used trial and error for this one too. Though I"m sure there is an easier way too.

Use the binomial theorem

$(a+n)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$

Since we have the $x^5$ term we know that

$n-k=5$

Also since we are on the 10th term we know k=9 (remember the index starts at 0 not 1) so we get

$n-9=5 \iff n=14$

See if you can figure out the next one with this theorem.

Good luck

3. Originally Posted by TheEmptySet
Use the binomial theorem

$(a+n)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$

Since we have the $x^5$ term we know that

$n-k=5$

Also since we are on the 10th term we know k=9 (remember the index starts at 0 not 1) so we get

$n-9=5 \iff n=14$

See if you can figure out the next one with this theorem.

Good luck