2025 AMC 10A Problems/Problem 19: Difference between revisions
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(Problem goes here) | |||
==Solution 1== | |||
Consider the polynomial <imath>f(x) = -x^2+3x+1.</imath> When we multiply this polynomial by <imath>x+1,</imath> we are essentially doing the operation given in the problem (When we multiply <imath>p(x)</imath> by <imath>x+1,</imath> a term of degree <imath>d</imath> in the yielded expression is the sum of <imath>1\cdot(\text{degree d})</imath> and <imath>x\cdot(\text{degree d-1})</imath> in <imath>p(x)</imath> This effect is visible in Pascal's Triangle). | |||
So, if we let the coefficients of <imath>f(x)</imath> be the zero row of the array, then the <imath>n^{th}</imath> row is just the coefficients of <imath>f(x)(x+1)^n.</imath> | |||
The next thing to note is that the sum of the coefficients in any polynomial <imath>p(x)</imath> is just <imath>p(1).</imath> Therefore, the sum of the entries in the <imath>n^{th}</imath> row of the array is <imath>f(1)(1+1)^n=3*2^n.</imath> Letting this equal <imath>12288,</imath> we get <imath>n=12.</imath> We are looking for the <imath>3^{rd}</imath> term in the <imath>12^{th}</imath> row. | |||
The <imath>12^{th}</imath> row is given by the coefficients of <imath>f(x)(x+1)^{12}=(-x^2+3x+1)(x+1)^{12}.</imath> Since the degree of the resulting expression is <imath>14,</imath> the third term in the row is just the coefficient of <imath>x^{12}</imath> in the expression, which is <imath>-\dbinom{12}{10}+3\dbinom{12}{11}+1=\boxed{\text{() }-29}.</imath> | |||
~Tacos_are_yummy_1 | |||
Revision as of 12:59, 6 November 2025
(Problem goes here)
Solution 1
Consider the polynomial 
When we multiply this polynomial by 
we are essentially doing the operation given in the problem (When we multiply 
by a term of degree 
in the yielded expression is the sum of 
and 
in This effect is visible in Pascal's Triangle).
So, if we let the coefficients of 
be the zero row of the array, then the 
row is just the coefficients of 
The next thing to note is that the sum of the coefficients in any polynomial is just 
Therefore, the sum of the entries in the row of the array is 
Letting this equal 
we get 
We are looking for the 
term in the 
row.
The row is given by the coefficients of 
Since the degree of the resulting expression is 
the third term in the row is just the coefficient of 
in the expression, which is 
~Tacos_are_yummy_1
