Difference between revisions of "2008 AMC 12A Problems/Problem 19"
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<math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math> | <math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math> | ||
− | ==Solution 1== | + | ==Solution 1 (easiest)== |
Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. | Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. | ||
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We determine the <math>x^{28}</math> coefficient by doing casework on the first three terms in our product. We can obtain an <math>x^{28}</math> term by choosing <math>x^{28}</math> in the first term, <math>-1</math> in the second and third terms, and <math>1</math> in the fourth term. We can get two <math>x^{28}</math> terms by choosing <math>x^{15}</math> in either the second or third term, <math>-1</math> in the first term, <math>-1</math> in the second or third term from which <math>x^{15}</math> has not been chosen, and the <math>\binom{15}{2}x^{13}</math> in the fourth term. We get <math>\binom{15}{2} * 2 = 210</math> <math>x^{28}</math> terms this way. (We multiply by <math>2</math> because the <math>x^{15}</math> term could have been chosen from the second term or the third term). Lastly, we can get an <math>x^{28}</math> term by choosing <math>-1</math> in the first three terms and a <math>\binom{30}{2}x^{28}</math> from the fourth term. We have a total of <math>1 + 210 - 435 = -224</math> for the <math>x^{28}</math> coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of <math>224 \Rightarrow \boxed{(C)}</math>. | We determine the <math>x^{28}</math> coefficient by doing casework on the first three terms in our product. We can obtain an <math>x^{28}</math> term by choosing <math>x^{28}</math> in the first term, <math>-1</math> in the second and third terms, and <math>1</math> in the fourth term. We can get two <math>x^{28}</math> terms by choosing <math>x^{15}</math> in either the second or third term, <math>-1</math> in the first term, <math>-1</math> in the second or third term from which <math>x^{15}</math> has not been chosen, and the <math>\binom{15}{2}x^{13}</math> in the fourth term. We get <math>\binom{15}{2} * 2 = 210</math> <math>x^{28}</math> terms this way. (We multiply by <math>2</math> because the <math>x^{15}</math> term could have been chosen from the second term or the third term). Lastly, we can get an <math>x^{28}</math> term by choosing <math>-1</math> in the first three terms and a <math>\binom{30}{2}x^{28}</math> from the fourth term. We have a total of <math>1 + 210 - 435 = -224</math> for the <math>x^{28}</math> coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of <math>224 \Rightarrow \boxed{(C)}</math>. | ||
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==See Also== | ==See Also== | ||
Latest revision as of 22:54, 9 January 2021
Problem
In the expansion of what is the coefficient of ?
Solution 1 (easiest)
Let and . We are expanding .
Since there are terms in , there are ways to choose one term from each . The product of the selected terms is for some integer between and inclusive. For each , there is one and only one in . For example, if I choose from , then there is exactly one power of in that I can choose; in this case, it would be . Since there is only one way to choose one term from each to get a product of , there are ways to choose one term from each and one term from to get a product of . Thus the coefficient of the term is .
Solution 2
Let . Then the term from the product in question is
So we are trying to find the sum of the coefficients of minus . Since the constant term in (when expanded) is , and the sum of the coefficients of is , we find the answer to be .
Solution 3
We expand to and use FOIL to multiply. It expands out to:
It becomes apparent that
.
Now we have to find the coefficient of in the product:
.
We quickly see that the we get terms from , , , ... , ... . The coefficient of is just the sum of the coefficients of all these terms. , so the answer is .
Solution 4
Rewrite the product as . It is known that
Thus, our product becomes
We determine the coefficient by doing casework on the first three terms in our product. We can obtain an term by choosing in the first term, in the second and third terms, and in the fourth term. We can get two terms by choosing in either the second or third term, in the first term, in the second or third term from which has not been chosen, and the in the fourth term. We get terms this way. (We multiply by because the term could have been chosen from the second term or the third term). Lastly, we can get an term by choosing in the first three terms and a from the fourth term. We have a total of for the coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of .
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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