If 3rd, 4th 5th and 6th terms in the expansion of (x + a)ⁿ be respectively a, b, c and d, prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}$ .

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Solution

$We have : (x + a)^{n} The 3 rd, 4 th, 5 th and 6 th terms are {}^{n}C_{2} x^{n - 2} a^{2}, {}^{n}C_{3} x^{n - 3} a^{3}, {}^{n}C_{4} x^{n - 4} a^{4} and {}^{n}C_{5} x^{n - 5} a^{5}, respectively . Now, {}^{n}C_{2} x^{n - 2} a^{2} = a {}^{n}C_{3} x^{n - 3} a^{3} = b {}^{n}C_{4} x^{n - 4} a^{4} = c {}^{n}C_{5} x^{n - 5} a^{5} = d LHS = \frac{b^{2} - a c}{c^{2} - b d}$

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If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that b2-acc2-bd=5a3c.

We have:(x+a)nThe 3rd, 4th, 5th and 6th terms are C2nxn-2 a2, C3nxn-3 a3, C4nxn-4 a4 and C5nxn-5 a5, respectively.Now,C2nxn-2 a2=aC3nxn-3 a3 =bC4nxn-4 a4 =cC5nxn-5 a5 =dLHS =b2-acc2-bd

If 3rd, 4th 5th and 6th terms in the expansion of (x + a)ⁿ be respectively a, b, c and d, prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}$ .