Difference between revisions of "1986 AIME Problems/Problem 11"

Revision as of 23:54, 3 March 2020

Problem

The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants. Find the value of $a_2$ .

Solution

Solution 1

Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the Binomial Theorem, this is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$ .

Solution 2

Again, notice $x = y - 1$ . So

$\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.$

We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$ . The Hockey Stick Identity tells us that this quantity is equal to ${18\choose 3} = \boxed{816}$ .

Solution 3

Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: $\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.$ From binomial theorem, the coefficient of the $y^2$ term is ${2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}$ . This is actually the sum of the first 16 triangular numbers, which evaluates to $\frac{(16)(17)(18)}{6} = \boxed{816}$ .

Solution 4(calculus)

Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ .

Then, since $f(x)=g(y)$ , $\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}$ $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.

Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})$

Now, notice that if $x = -1$ , then $y = 0$ , so $f^{''}(-1) = g^{''}(0)$

$g^{''}(0)= 2a_2$ , and $f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17$ .

Now, we can use the hockey stick theorem to see that $2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}$

Thus, $2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}$

-AOPS81619

@@ Line 36: / Line 36: @@
 Now, we can use the hockey stick theorem to see that <math>2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}</math>
-Thus, <math>2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=</math>
+Thus, <math>2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}</math>
+-AOPS81619
 == See also ==

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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