Question

Find the term in the expansion of $(2x - 5)^6$ which have

(i) Greatest binomial coefficient $\rightarrow$ middle term (ii) Greatest numerical coefficient (iii) Algebrically greatest coefficient (iv) Algebrically least coefficient

Answer 1

Greatest binomial coefficient: $T_4 = -20000x^3$

Greatest numerical coefficient: $T_5 = 37500x^2$ and $T_6 = -37500x$

Algebraically greatest coefficient: $T_5 = 37500x^2$

Algebraically least coefficient: $T_6 = -37500x$

Answer 2

The binomial expansion of $(2x - 5)^6$ is given by the formula $(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$. Here, $a = 2x$, $b = -5$, and $n = 6$. The general term (r+1)th term is $T_{r+1} = \binom{6}{r} (2x)^{6-r} (-5)^r = \binom{6}{r} 2^{6-r} x^{6-r} (-5)^r$. The coefficient of the term $T_{r+1}$ is $C_r = \binom{6}{r} 2^{6-r} (-5)^r$.

(i) Greatest binomial coefficient: For the expansion of $(a+b)^n$, the greatest binomial coefficient occurs in the middle term(s). Since $n=6$ is even, there are $n+1=7$ terms, and the middle term is the $(6/2 + 1) = 4$th term. The binomial coefficient of the 4th term ($r=3$) is $\binom{6}{3}$. $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. The term with the greatest binomial coefficient is $T_4$. $T_4 = \binom{6}{3} (2x)^{6-3} (-5)^3 = 20 (2x)^3 (-5)^3 = 20 (8x^3) (-125) = 160x^3 (-125) = -20000x^3$. The term with the greatest binomial coefficient is $-20000x^3$.

(ii) Greatest numerical coefficient: The numerical coefficient of the term $T_{r+1}$ is $|C_r| = |\binom{6}{r} 2^{6-r} (-5)^r| = \binom{6}{r} 2^{6-r} |-5|^r = \binom{6}{r} 2^{6-r} 5^r$. Let $N_r = \binom{6}{r} 2^{6-r} 5^r$. We need to find the value of $r$ (from 0 to 6) for which $N_r$ is maximum. Consider the ratio $\frac{N_{r+1}}{N_r} = \frac{\binom{6}{r+1} 2^{5-r} 5^{r+1}}{\binom{6}{r} 2^{6-r} 5^r} = \frac{6-r}{r+1} \frac{5}{2}$. The numerical coefficient increases when $\frac{N_{r+1}}{N_r} > 1$. $\frac{6-r}{r+1} \frac{5}{2} > 1 \implies 5(6-r) > 2(r+1) \implies 30 - 5r > 2r + 2 \implies 28 > 7r \implies 4 > r$. So, $N_0 < N_1 < N_2 < N_3 < N_4$. The numerical coefficient decreases when $\frac{N_{r+1}}{N_r} < 1$. $\frac{6-r}{r+1} \frac{5}{2} < 1 \implies 4 < r$. So, $N_5 > N_6$ (for $r=5$). When $r=4$, $\frac{N_5}{N_4} = \frac{6-4}{4+1} \frac{5}{2} = \frac{2}{5} \frac{5}{2} = 1$. So $N_5 = N_4$. The sequence of numerical coefficients is $N_0 < N_1 < N_2 < N_3 < N_4 = N_5 > N_6$. The greatest numerical coefficient is $N_4 = N_5$. $N_4 = \binom{6}{4} 2^{6-4} 5^4 = 15 \times 2^2 \times 5^4 = 15 \times 4 \times 625 = 60 \times 625 = 37500$. $N_5 = \binom{6}{5} 2^{6-5} 5^5 = 6 \times 2^1 \times 5^5 = 12 \times 3125 = 37500$. The terms with the greatest numerical coefficient are $T_5$ ($r=4$) and $T_6$ ($r=5$). $T_5 = \binom{6}{4} (2x)^2 (-5)^4 = 15 (4x^2) (625) = 60x^2 (625) = 37500x^2$. $T_6 = \binom{6}{5} (2x)^1 (-5)^5 = 6 (2x) (-3125) = 12x (-3125) = -37500x$. The terms with the greatest numerical coefficient are $37500x^2$ and $-37500x$.

(iii) Algebraically greatest coefficient: The coefficients are $C_r = \binom{6}{r} 2^{6-r} (-5)^r$. $C_0 = \binom{6}{0} 2^6 (-5)^0 = 1 \times 64 \times 1 = 64$. $C_1 = \binom{6}{1} 2^5 (-5)^1 = 6 \times 32 \times (-5) = 6 \times (-160) = -960$. $C_2 = \binom{6}{2} 2^4 (-5)^2 = 15 \times 16 \times 25 = 15 \times 400 = 6000$. $C_3 = \binom{6}{3} 2^3 (-5)^3 = 20 \times 8 \times (-125) = 20 \times (-1000) = -20000$. $C_4 = \binom{6}{4} 2^2 (-5)^4 = 15 \times 4 \times 625 = 60 \times 625 = 37500$. $C_5 = \binom{6}{5} 2^1 (-5)^5 = 6 \times 2 \times (-3125) = 12 \times (-3125) = -37500$. $C_6 = \binom{6}{6} 2^0 (-5)^6 = 1 \times 1 \times 15625 = 15625$. The coefficients are 64, -960, 6000, -20000, 37500, -37500, 15625. The algebraically greatest coefficient is the maximum value in this list, which is 37500. This coefficient corresponds to $r=4$, which is the term $T_5$. $T_5 = 37500x^2$. The term with the algebraically greatest coefficient is $37500x^2$.

(iv) Algebraically least coefficient: The algebraically least coefficient is the minimum value in the list of coefficients: 64, -960, 6000, -20000, 37500, -37500, 15625. The minimum value is -37500. This coefficient corresponds to $r=5$, which is the term $T_6$. $T_6 = -37500x$. The term with the algebraically least coefficient is $-37500x$.

Summary of the terms: (i) Greatest binomial coefficient: $T_4 = -20000x^3$. (ii) Greatest numerical coefficient: $T_5 = 37500x^2$ and $T_6 = -37500x$. (iii) Algebraically greatest coefficient: $T_5 = 37500x^2$. (iv) Algebraically least coefficient: $T_6 = -37500x$.