InterviewSolution
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InterviewSolution
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If the 3rd, 4th , 5th and 6th term in the expansion of`(x+alpha)^n`be, respectively, `a ,b ,ca n dd ,`prove that `(b^2-a c)/(c^2-b d)=(5a)/(3c)dot` |
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Answer» We know that `(T_(r+1))/(T_(r)), (n-r+1)/(r)(alpha)/(x)` `T_(3)=a,T_(4)=b,T_(5)=c,T_(6)=d` Putting `r = 3, 4,5` in the abvoe we get `(T_(4))/(T_(3)) = (n-2)/(3)(alpha)/(x) = b/a` `(T_(5))/(T_(4)) = (n-3)/(4)(alpha)/(x) = c/b` `(T_(6))/(T_(5)) = (n-4)/(5) (alpha)/(x) = d/c` We have to prove `(b^(2)-ac)/(c^(2)-bd) = (5a)/(3c)` or `((b)/(c)-(a)/(b))/((c)/(b)-(d)/(c)) = (5a)/(3c)` Now, `((b)/(c) - (a)/(b))/((c)/(b)-(d)/(c)) = ((4x)/((n-3)alpha)-(3x)/((n-2)alpha))/(((n-3)alpha)/(4x) - ((n-4)alpha)/(5x))` `= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))(4n-8-3n+9)/(5n-15-4n+16)` `= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))` Also, `(5a)/(3c) = (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))` `:. L.H.S. = R.H.S.` |
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