How Do You Find the Coefficient \( a_2 \) in a Polynomial Transformation?

[anemone]

Here is this week's POTW:

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The polynomial $1-y+y^2-y^3+\cdots+y^{16}-y^{17}$

may be written in the form $a_0+a_1x+a_2x^2+\cdots+a_{16}x^{16}+a_{17}x^{17}$, where $x=y+1$ and $a_i$ are constants.

Find $a_2$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. lfdahl

Solution from MarkFL:

Spoiler

Let:

\(\displaystyle f(y)=\sum_{k=0}^{17}\left((-y)^k\right)\)

Hence, using $-y=1-x$, we may write:

\(\displaystyle f(x)=\sum_{k=0}^{17}\left((1-x)^k\right)=\frac{1}{x}\left(1-(x-1)^{18}\right)\)

Now, we may observe that this may be written in the form:

\(\displaystyle f(x)=\sum_{k=0}^{17}\left(a_{k}x^k\right)\)

where:

\(\displaystyle a_k=(-1)^k{18 \choose k+1}\)

Hence:

\(\displaystyle a_2={18 \choose 3}=816\)

Alternate solution from lfdahl:

Spoiler

\[1-y+y^2-y^3+ ...+y^{16}-y^{17} = \sum_{k=0}^{17}(-1)^ky^k=\sum_{k=0}^{17}(-1)^k(x-1)^k=\sum_{k=0}^{17}(1-x)^k \\\\ =\sum_{k=0}^{17}\sum_{i=0}^{k}\binom{k}{i}(-x)^i\]

Counting the number of $x^2$ using Pascals triangle formula (for $i=2$ and $k \ge 2$):

\[a_2 = \binom{2}{2}+\binom{3}{2}+...+\binom{17}{2} = \sum_{j=2}^{17}\binom{j}{2} \\\\= 1+3+6+10+15+...+120+136 \\\\= 1 + (1+2)+(1+2+3)+ ...+(1+2+3+...15+16) \\\\= 1 + (2+1)\frac{2}{2}+(3+1)\frac{3}{2}+(4+1)\frac{4}{2}+...+(16+1)\frac{16}{2} \\\\= \frac{1}{2}\sum_{k=1}^{16}(k+1)k \\\\= \frac{1}{2}\sum_{k=1}^{16}k^2+\frac{1}{2}\sum_{k=1}^{16}k \\\\= \frac{1}{2}\frac{16(16+1)(32+1)}{6}+\frac{1}{2}(16+1)\frac{16}{2} \\\\= 816.\]

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. greg1313

Solution from MarkFL:

Spoiler

Let:

\(\displaystyle f(y)=\sum_{k=0}^{17}\left((-y)^k\right)\)

Hence, using $-y=1-x$, we may write:

\(\displaystyle f(x)=\sum_{k=0}^{17}\left((1-x)^k\right)=\frac{1}{x}\left(1-(x-1)^{18}\right)\)

Now, we may observe that this may be written in the form:

\(\displaystyle f(x)=\sum_{k=0}^{17}\left(a_{k}x^k\right)\)

where:

\(\displaystyle a_k=(-1)^k{18 \choose k+1}\)

Hence:

\(\displaystyle a_2={18 \choose 3}=816\)

Alternate solution from greg1313:

Spoiler

\[1-y+y^2-y^3+ ...+y^{16}-y^{17} = \sum_{k=0}^{17}(-1)^ky^k=\sum_{k=0}^{17}(-1)^k(x-1)^k=\sum_{k=0}^{17}(1-x)^k \\\\ =\sum_{k=0}^{17}\sum_{i=0}^{k}\binom{k}{i}(-x)^i\]

Counting the number of $x^2$ using Pascals triangle formula (for $i=2$ and $k \ge 2$):

\[a_2 = \binom{2}{2}+\binom{3}{2}+...+\binom{17}{2} = \sum_{j=2}^{17}\binom{j}{2} \\\\= 1+3+6+10+15+...+120+136 \\\\= 1 + (1+2)+(1+2+3)+ ...+(1+2+3+...15+16) \\\\= 1 + (2+1)\frac{2}{2}+(3+1)\frac{3}{2}+(4+1)\frac{4}{2}+...+(16+1)\frac{16}{2} \\\\= \frac{1}{2}\sum_{k=1}^{16}(k+1)k \\\\= \frac{1}{2}\sum_{k=1}^{16}k^2+\frac{1}{2}\sum_{k=1}^{16}k \\\\= \frac{1}{2}\frac{16(16+1)(32+1)}{6}+\frac{1}{2}(16+1)\frac{16}{2} \\\\= 816.\]

Hi MHB,

I mistook lfdahl's submission of a solution to last week's Secondary School/High School POTW problem as greg1313's and therefore I made a slip in the declaration of the names of those who correctly answered last week's POTW (High School) problem.

In an effort to correct this egregious error, please accept my heartfelt apology, lfdahl, our POTW's ardent fan, and I would also like to extend a sincere thank you to another loyal member, greg1313 for your PM informing me about it.