106put n=1, a,b get eliminated
put n=2, (2b^2/a^2 - 2c/a) which is (d)
hence ans (d)
To seriously say,, i could not get the series..
let α & β be the roots of the equation ax^2+bx+c=0,a≠0,then the sum of the series (α+β)^(2 )+(α^2+β^2 )+(α-β)^2+⋯upto n terms is equal to
A)(n(a^2+(n-1)ac))/a^2 B)(n(a^2-(n-1)bc))/b^2 C)(n(b^2+(n-1)ac))/a^2
D)(n(b^2-(n-1)ac))/a^2
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2 Answers
106
29@Asish the series is α2 + β2 + 2αβ , α2 + β2 , α2 + β2 - 2αβ, α2 + β2 - 4αβ ,....
the answers is D however..
solve For sum of n terms of an AP with common difference -2αβ
