" 6."(ax)^(n)+b^(n)

(ax)^(m)+(b)^(n)

If one root is nth power of the other root of this equation x^(2)-ax+b=0 then, b^(n/(n+1))+b^(1/(n+1)) = (A) a (B) a^(n) (C) b^(n) (D) ab

Integral of the form: (ax+b)^(n)P(x)dx;P(x)/((ax+b)^(n))dx

A sequence , a , ax , ax^(2), "……" ax^(n) , has odd number of terms . Find its median . (a) ax^(n-1) (b) ax^((n//2) - 1) (c) ax^(n//2) (d) ax^((n//2) + 1)

If n is any natural number,then 6^(n)-5^(n) always ends with (a) 1 (b) 3 (c) 5 (d) 7 [Hint: For any n in N,6^(n) and 5^(n) end with 6 and 5 respectively.Therefore,6^(n)-5^(n) always ends with 6-5=1.

If y=ax^(n+1)+bx^(-n), then x^(2)(d^(2)y)/(dx^(2)) is equal to n(n-1)y(b)n(n+1)yny(d)n^(2)y

If y=ax^(n+1)+bx^(-n), then x^(2)(d^(2)y)/(dx^(2))=n(n-1)y(b)n(n+1)y(c)ny(d)n^(2)y

By substitution: Theorem: If int(ax+b)^(n)dx=((ax+b)^(n+1))/(a)(n+1)

If (1 + ax)^(n) = 1 + 6x + 27/2 x^(2) + ...... + a^(n)x^(n) , then the values of a and n are respectively