If in any binomial expansion a, b, c and d be the 6th, 7th, 8th and 9th terms respectively, prove that `(b^2-ac)/(c^2-bd)=(4a)/(3c)`

Explanation

Suppose expression is `(1+x)^{n}`.


Then, the 6 th, 7 th, 8 th and 9 th terms are `\ ^{n} C_{5} x^{5},\ ^{n} C_{6} x^{6},\ ^{n} C_{7} x^{7}` and `\ ^{n} C_{8} x^{8}`, respectively .


Now, we have:


`frac{\ ^{n} C_{6 x^{5}}}{\ ^{n} C_{5} x^{5}}=frac{b}{a}, frac{\ ^{n} C_{8 x^{8}}}{\ ^{n} C_{7} x^{7}}=frac{d}{c} and frac{\ ^{n} C_{7 x^{7}}}{\ ^{n} C_{6} x^{6}}=frac{c}{b} `


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