Binomial Theorem flash cards

1. 1.State the general term $T_{r+1}$ in the expansion of $(x+y)^n$.

   $T_{r+1} = {}^nC_r\, x^{n-r} y^r$, where $r=0,1,2,\dots,n$.

   Hint: Binomial coefficient times decreasing power of $x$.

2. 2.In $(x+y)^n$, how do you find the term independent of $x$ (constant term)?

   Write $T_{r+1}={}^nC_r\, x^{n-r} y^r$, set the net power of $x$ equal to $0$, solve for $r$, then substitute that value of $r$ back into $T_{r+1}$ to get the constant term.

   Hint: Combine exponents from both parts and equate to zero.

3. 3.For which value(s) of $r$ is the binomial coefficient ${}^nC_r$ in $(1+x)^n$ greatest?

   If $n$ is even, ${}^nC_r$ is greatest at $r=\dfrac{n}{2}$; if $n$ is odd, it is greatest at $r=\dfrac{n-1}{2}$ and $r=\dfrac{n+1}{2}$ (the two values give equal, maximum coefficients).

   Hint: Middle term(s) of the expansion.

4. 4.How do you determine the numerically greatest term in the expansion of $(1+x)^n$?

   Form the ratio $\left|\dfrac{T_{r+1}}{T_r}\right|=\dfrac{n-r+1}{r}|x|$ and find the value of $r$ at which this ratio changes from $\ge 1$ (term still increasing) to $<1$ (term starts decreasing); that $T_{r+1}$ is the numerically greatest term. If the ratio equals exactly $1$ at some integer $r$, then $T_r$ and $T_{r+1}$ are equal and both are the greatest.

   Hint: Use the consecutive-term ratio test.

5. 5.What are the standard sums $\sum {}^nC_r$ and $\sum (-1)^r\,{}^nC_r$ for $r=0$ to $n$?

   $\displaystyle\sum_{r=0}^{n} {}^nC_r = 2^n$ and $\displaystyle\sum_{r=0}^{n} (-1)^r\,{}^nC_r = 0$ (for $n\ge 1$).

   Hint: Put $x=1$ and $x=-1$ in $(1+x)^n$.

6. 6.Find the coefficient of $x^5$ in the expansion of $\left(x^2-\dfrac{1}{x^3}\right)^{15}$.

   General term $T_{r+1}=\binom{15}{r}(x^2)^{15-r}\left(-\dfrac{1}{x^3}\right)^r=(-1)^r\binom{15}{r}x^{30-5r}$. Setting $30-5r=5$ gives $r=5$. So the coefficient is $(-1)^5\binom{15}{5}=-3003$.

   Hint: Write the general term and equate the power of $x$ to 5.

7. 7.Find the term independent of $x$ in the expansion of $\left(\dfrac{3}{2}x^2-\dfrac{1}{3x}\right)^9$.

   General term $T_{r+1}=\binom{9}{r}\left(\dfrac{3}{2}\right)^{9-r}x^{2(9-r)}\left(-\dfrac{1}{3x}\right)^r$, whose power of $x$ is $18-3r$; setting this to $0$ gives $r=6$. The term equals $\binom{9}{6}\left(\dfrac{3}{2}\right)^3\left(\dfrac{1}{3}\right)^6=84\cdot\dfrac{27}{8}\cdot\dfrac{1}{729}=\dfrac{7}{18}$.

   Hint: Set the exponent of $x$ in the general term to zero.

8. 8.If the coefficients of the 5th, 6th and 7th terms in the expansion of $(1+x)^n$ are in Arithmetic Progression, find all possible values of $n$.

   The condition gives $\binom{n}{4},\binom{n}{5},\binom{n}{6}$ in AP, so $2\binom{n}{5}=\binom{n}{4}+\binom{n}{6}$. Dividing through by $\binom{n}{5}$ and using $\binom{n}{4}/\binom{n}{5}=5/(n-4)$, $\binom{n}{6}/\binom{n}{5}=(n-5)/6$ leads to $n^2-21n+98=0$, giving $n=7$ or $n=14$.

   Hint: Convert the AP condition into a ratio equation using $\binom{n}{r}/\binom{n}{r-1}$.

9. 9.Find the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$.

   Write $(1+x)(1-x)^n=(1-x)^n+x(1-x)^n$. The coefficient of $x^n$ from the first term is $(-1)^n$, and from the second term equals the coefficient of $x^{n-1}$ in $(1-x)^n$, which is $(-1)^{n-1}n$. Adding, the coefficient of $x^n$ is $(-1)^n+(-1)^{n-1}n=(-1)^{n-1}(n-1)$.

   Hint: Split the product into two separate expansions before comparing coefficients.

10. 10.In the expansion of $\left(x^2+\dfrac{2}{x}\right)^{15}$, find the ratio of the coefficient of $x^{15}$ to the term independent of $x$.

    General term $T_{r+1}=\binom{15}{r}2^r x^{30-3r}$. For $x^{15}$: $r=5$, coefficient $=\binom{15}{5}2^5$. For the constant term: $r=10$, coefficient $=\binom{15}{10}2^{10}$. Since $\binom{15}{5}=\binom{15}{10}$, the ratio is $2^5/2^{10}=\dfrac{1}{32}$.

    Hint: Use the symmetry $\binom{15}{5}=\binom{15}{10}$ to simplify the ratio.

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