General and Middle Terms, with Mixed Practice

Exam­ple 1. Find the mid­dle term of $\displaystyle \left(x - \tfrac{2}{x}\right)^8$: $\displaystyle T_5 = {}^8C_4\, x^4 \left(-\tfrac{2}{x}\right)^4 = 70 \times 16 = 1120$.

Exam­ple 2. Now the mid­dle terms of $(2x + y)^7$: $T_4 = {}^7C_3 (2x)^4 y^3 = 560x^4y^3$ and $T_5 = {}^7C_4 (2x)^3 y^4 = 280x^3y^4$.

Exam­ple 3. We want the coef­fi­cient of $x^5$ in $(x + 3)^8$: $T_{r+1} = {}^8C_r x^{8-r} 3^r$. The pow­ers must match, so $8 - r = 5$, giv­ing $r = 3$, and the coef­fi­cient is $^8C_3 \cdot 27 = 1512$.

Exam­ple 4. Look for the term inde­pen­dent of $x$ in $\displaystyle \left(x^2 - \tfrac{1}{x}\right)^9$: $T_{r+1} = {}^9C_r x^{18-2r}(-1)^r x^{-r}$. Set the total power to zero: $18 - 3r = 0$, so $r = 6$, and the term is $^9C_6 = 84$.

Exam­ple 5. The coef­fi­cient of $x^6y^3$ in $(x + 2y)^9$ comes straight from the gen­eral term: $^9C_3 \cdot 2^3 = 672$.

Exam­ple 6. If the coef­fi­cients of $x^{r-1}$, $x^r$, $x^{r+1}$ in $(1 + x)^n$ are in the ratio $1 : 3 : 5$, com­pare neigh­bour­ing coef­fi­cients: $\displaystyle \tfrac{^nC_r}{^nC_{r-1}} = \tfrac{n - r + 1}{r} = 3$ and $\displaystyle \tfrac{^nC_{r+1}}{^nC_r} = \tfrac{n - r}{r + 1} = \tfrac{5}{3}$. Solve the pair and you get $n = 7$, $r = 2$.

Exam­ple 7. If the $3$rd, $4$th and $5$th terms of $(a + b)^n$ are $84$, $280$ and $560$, divide one term by the one before it: $\displaystyle \tfrac{T_4}{T_3} = \tfrac{n - 2}{3}\cdot\tfrac{b}{a} = \tfrac{10}{3}$ and $\displaystyle \tfrac{T_5}{T_4} = \tfrac{n - 3}{4}\cdot\tfrac{b}{a} = 2$. Divid­ing these two gives $\displaystyle \tfrac{4(n - 2)}{3(n - 3)} = \tfrac{5}{3}$, $n = 7$, $\displaystyle \tfrac{b}{a} = 2$. Finally, $T_3 = 21a^5b^2 = 84a^7 = 84$ gives $a = 1$, $b = 2$.