If 3rd,4th,5th,6th term in the expansion of (x+alpha)n be respectively a,b,c and d, prove that b2-ac/c2-bd=4a/3c - Maths - Binomial Theorem

Question

If 3rd,4th,5th,6th term in the expansion of
(x+alpha)n be respectively a,b,c and d, prove that
b2-ac/c2-bd=4a/3c

Ankush Jain · Accepted Answer

Given expansion is: $(x + α)^{n}$ Then, $T_{3} = a = {}^{n}{C_{2} x^{n - 2} α^{2} = \frac{n (n - 1)}{2} x^{n - 2} α^{2}}$ $T_{4} = b = {}^{n}{C_{3} x^{n - 3} α^{3} = \frac{n (n - 1) (n - 2)}{6} x^{n - 3} α^{3}}$ $T_{5} = c = {}^{n}{C_{4} x^{n - 4} α^{4} = \frac{n (n - 1) (n - 2) (n - 3)}{24} x^{n - 4} α^{4}}$ $T_{6} = d = {}^{n}{C_{5} x^{n - 5} α^{5} = \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{120} x^{n - 5} α^{5}}$ Now, $\frac{T_{4}}{T_{3}} = \frac{b}{a} = \frac{\frac{n (n - 1) (n - 2)}{6} x^{n - 3} α^{3}}{\frac{n (n - 1)}{2} x^{n - 2} α^{2}} = (\frac{n - 2}{3}) \frac{α}{x} (1)$ Similarly, $\frac{T_{5}}{T_{4}} = \frac{c}{b} = \frac{\frac{n (n - 1) (n - 2) (n - 3)}{24} x^{n - 4} α^{4}}{\frac{n (n - 1) (n - 2)}{6} x^{n - 3} α^{3}} = (\frac{n - 3}{4}) \frac{α}{x} (2) a n d \frac{T_{6}}{T_{5}} = \frac{d}{c} = \frac{\frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{120} x^{n - 5} α^{5}}{\frac{n (n - 1) (n - 2) (n - 3)}{24} x^{n - 4} α^{4}} = (\frac{n - 4}{5}) \frac{α}{x} (3)$ Again, dividing (1) by (2) and (2) by (3), we get $\frac{\frac{b}{a}}{\frac{c}{b}} = \frac{(\frac{n - 2}{3}) \frac{α}{x}}{(\frac{n - 3}{4}) \frac{α}{x}} = \frac{4 (n - 2)}{3 (n - 3)} \Rightarrow \frac{b^{2}}{a c} = \frac{4 (n - 2)}{3 (n - 3)} (4) a n d \frac{\frac{c}{b}}{\frac{d}{c}} = \frac{(\frac{n - 3}{4}) \frac{α}{x}}{(\frac{n - 4}{5}) \frac{α}{x}} = \frac{5 (n - 3)}{4 (n - 4)} \Rightarrow \frac{c^{2}}{b d} = \frac{5 (n - 3)}{4 (n - 4)} (5)$ Now subtracting 1 from both sides of equation (4) and (5) as: $\frac{b^{2}}{a c} - 1 = \frac{4 (n - 2)}{3 (n - 3)} - 1 \Rightarrow \frac{b^{2} - a c}{a c} = \frac{n + 1}{3 (n - 3)} (6) a n d \frac{c^{2}}{b d} - 1 = \frac{5 (n - 3)}{4 (n - 4)} - 1 \Rightarrow \frac{c^{2} - b d}{b d} = \frac{n + 1}{4 (n - 4)} (7)$ Again, on dividing (6) by (7), we get $\frac{\frac{b^{2} - a c}{a c}}{\frac{c^{2} - b d}{b d}} = \frac{\frac{n + 1}{3 (n - 3)}}{\frac{n + 1}{4 (n - 4)}} \Rightarrow \frac{b^{2} - a c}{c^{2} - b d} \frac{b d}{a c} = \frac{4 (n - 4)}{3 (n - 3)} (8)$ On multiplying (5) by (8), we get $\Rightarrow \frac{b^{2} - a c}{c^{2} - b d} \frac{b d}{a c} \times \frac{c^{2}}{b d} = \frac{4 (n - 4)}{3 (n - 3)} \times \frac{5 (n - 3)}{4 (n - 4)} \Rightarrow \frac{b^{2} - a c}{c^{2} - b d} \frac{c}{a} = \frac{5}{3} \Rightarrow \frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c} (9)$ The RHS of equation (9) is 5a/3c not 4a/3c Please recheck it

If 3rd,4th,5th,6th term in the expansion of (x+alpha)n be respectively a,b,c and d, prove that b2-ac/c2-bd=4a/3c..

If 3rd,4th,5th,6th term in the expansion of (x+alpha)ⁿ be respectively a,b,c and d, prove that b²-ac/c²-bd=4a/3c..