ANOTHER Binomial Q (1 Viewer)

iSplicer · Dec 18, 2009

Hey guys, just another quick one here =)

in the expansion of (1+3x+ax^2)^n, where n is a positive integer, the coefficient of x^2 is 0. Find, in terms of n, the value of a and the coefficient of x^3

Thanks!

Trebla · Dec 18, 2009

iSplicer said:
Hey guys, just another quick one here =)

in the expansion of (1+3x+ax^2)^n, where n is a positive integer, the coefficient of x^2 is 0. Find, in terms of n, the value of a and the coefficient of x^3

Thanks!

$(1+3x+ax^2)^n=\sum_{r=0}^{n}\binom{n}{r}(3x+ax^2)^{r}(1)^{n-r}\\\\=\sum_{r=0}^{n}\binom{n}{r}x^r(3+ax)^{r}\\\\$Clearly, the term with $x^2$ cannot occur if $r$ is higher than $2$. \\When $r=0$, there is clearly no $x^2$ term. So consider $r=1,2\\\\r=1$ the term is $\binom{n}{1}x(3+ax)$. So $x^2$ term by inspection is $ \binom{n}{1}ax^2\\\\r=2$ the term is $\binom{n}{2}x^2(3+ax)^2$. So $x^2$ term by inspection is $ \binom{n}{2}9x^2\\\\$Hence the coefficient of $x^2$ is $\binom{n}{1}a + \binom{n}{2}9=na+\dfrac{9n!}{2!(n-2)!}\\\\$This simplifies to $na+\dfrac{9n(n-1)}{2}\\\\$But coefficient of $x^2$ equals zero hence $\\\\n\left (a+\dfrac{9(n-1)}{2} \right )=0\Rightarrow a=\dfrac{9(1-n)}{2}\\(n \neq 0 $ since it is a positive integer according to the question$)$

iSplicer · Dec 18, 2009

Trebla said:
$(1+3x+ax^2)^n=\sum_{r=0}^{n}\binom{n}{r}(3x+ax^2)^{r}(1)^{n-r}\\\\=\sum_{r=0}^{n}\binom{n}{r}x^r(3+ax)^{r}\\\\$Clearly, the term with $x^2$ cannot occur if $r$ is higher than $2$. \\When $r=0$, there is clearly no $x^2$ term. So consider $r=1,2\\\\r=1$ the term is $\binom{n}{1}x(3+ax)$. So $x^2$ term by inspection is $ \binom{n}{1}ax^2\\\\r=2$ the term is $\binom{n}{2}x^2(3+ax)^2$. So $x^2$ term by inspection is $ \binom{n}{2}9x^2\\\\$Hence the coefficient of $x^2$ is $\binom{n}{1}a + \binom{n}{2}9=na+\dfrac{n!}{2!(n-2)!}\\\\$This simplifies to $na+\dfrac{n(n-1)}{2}\\\\$But coefficient of $x^2$ equals zero hence $\\\\n\left (a+\dfrac{n-1}{2} \right )=0\Rightarrow a=\dfrac{1-n}{2}\\(n \neq 0 $ since it is a positive integer according to the question$)$

Your solution is beautiful, thanks so much! Oh, and you forgot to include the factor of '9' in your working... you had it in the fifth last line but you forgot to carry it over to the fourth last line =)

Trebla · Dec 18, 2009

$$Consider $\\\\\displaystyle\sum_{r=0}^{n}\binom{n}{r}x^r(3+ax)^{r}\\\\$Clearly, the term with $x^3$ cannot occur if $r$ is higher than $3$. \\When $r=0,1$, there is clearly no $x^3$ term. So consider $r=2,3\\\\r=2$ the term is $\binom{n}{2}x^2(3+ax)^2$. So $x^3$ term by inspection is $ \binom{n}{2}x^2(6ax)\\\\r=3$ the term is $\binom{n}{3}x^3(3+ax)^3$. So $x^3$ term by inspection is $ \binom{n}{3}x^33^3\\\\$Hence the coefficient of $x^3$ is $\binom{n}{2}6a + \binom{n}{3}27=6a\dfrac{n(n-1)}{2}+27\dfrac{n(n-1)(n-2)}{6}\\\\=3an(n-1)+\dfrac{9n(n-1)(n-2)}{2}\;\;$but $a=\dfrac{9(1-n)}{2}\\\\=-\dfrac{27n(n-1)^2}{2}+\dfrac{9n(n-1)(n-2)}{2}$

Looks messy...not sure if it's right...

addikaye03 · Dec 19, 2009

Trebla said:
$(1+3x+ax^2)^n=\sum_{r=0}^{n}\binom{n}{r}(3x+ax^2)^{r}(1)^{n-r}\\\\=\sum_{r=0}^{n}\binom{n}{r}x^r(3+ax)^{r}\\\\$Clearly, the term with $x^2$ cannot occur if $r$ is higher than $2$. \\When $r=0$, there is clearly no $x^2$ term. So consider $r=1,2\\\\r=1$ the term is $\binom{n}{1}x(3+ax)$. So $x^2$ term by inspection is $ \binom{n}{1}ax^2\\\\r=2$ the term is $\binom{n}{2}x^2(3+ax)^2$. So $x^2$ term by inspection is $ \binom{n}{2}9x^2\\\\$Hence the coefficient of $x^2$ is $\binom{n}{1}a + \binom{n}{2}9=na+\dfrac{9n!}{2!(n-2)!}\\\\$This simplifies to $na+\dfrac{9n(n-1)}{2}\\\\$But coefficient of $x^2$ equals zero hence $\\\\n\left (a+\dfrac{9(n-1)}{2} \right )=0\Rightarrow a=\dfrac{9(1-n)}{2}\\(n \neq 0 $ since it is a positive integer according to the question$)$

Very Beautiful

ANOTHER Binomial Q (1 Viewer)

iSplicer

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Trebla

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iSplicer

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Trebla

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addikaye03

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