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)`

Answer

N/a

Explanation

In the expansion of `(1+x)^(n)`
`(T_(r+1))/(T_(r))=(.^(n)C_(r).x^(r))/(.^(n)C_(r-1).x^(r-1))`
`=(|uln|ul(r-1)|ul(n-r+1))/(|ulr|ul(n-r)|uln).x`
`=(|ul(r-1)(n-r+1)|ul(n-r))/(r|ul(r-1)|ul(n-r)).x`
`rArr " "(T_(r+1))/(T_(r))=(n-r+1)/(r).x`
`:. (T_(7))/(T_(6))=(n-6+1)/(6).x`
`rArr " "(b)/(a)=(n-5)/(6).x " "....(1)`
`(T_(8))/(T_(7))=(n-7+1)/(7).x`
`rArr " "(C)/(b)=(n-6)/(7)x" "....(2)`
`" and "(T_(9))/(T_(8))=(n-8+1)/(8).x`
`rArr " "(d)/(C) =(n-7)/(8).x" "......(3)` ltbr. Divide eq. (1) from (2) and eq. (2) from (3)
`(b^(2))/(ac)=(7(n-5))/(6(n-6))" "....(4)`
`" and " (c^(2))/(bd)=(8(n-6))/(7(n-7))" "......(5)`
`"Now "(b^(2))/(ac)-1=(7n-35)/(6n-36)-1`
`rArr " "(b^(2)-ac)/(ac)=(1+n)/(6n-36)....(6)`
`" and " (c^(2))/(bd)-1=(8n-84)/(7n-49)-1`
`rArr " "(c^(2)-bd)/(bd)=(n+1)/(7n-49)" ".....(7)`
Divide eq (6) by eq. (7)
`(b^(2)-ac)/(ac).(bd)/(c^(2)-bd)=(7n-49)/(6n-36)`
`(b^(2)-ac)/(c^(2)-bd).(bd)/(ac)=(7(n-7))/(6(n-6))" "....(8)`
Divide eq. (3) by eq. (2)
`(dxxb)/(cxxc)=(n-7)/(8).(7)/(n-6)`
`rArr " "(8bd)/(7c^(2))=(n-7)/(n-6)" "......(9)`
From eqs (8) and (9)
`(b^(2)-ac)/(c^(2)-bd).(bd)/(ac)-(7)/(6).(8bd)/(7c^(2))`
`rArr " "(b^(2)-ac)/(c^(2)-bd)=(4a)/(3c)` Hence Proved.