[quad |a_(1)+2a_(2)+3a_(3)+.........+na_(n)|<=1],[" The function "f:R rarr R" satisfies "f(x^(2))*f''(x)=f'(x)*f'(x^(2))" for all real "x" .Given that "f(1)=1" and "],[f''(1)=8," compute the value of "f'(1)+f''(1)]

If |a_1Sinx+a_(2)sin2x+.....+a_(n)sinx|le|sinx| for x in R then maimum value of |a_(1)+2a_(2)+3a_(3)+.....+na_(n)| is

If a_(1),a_(2) , a_(3),"……..",a_(n) are in H.P. , then : (a_(1))/( a_(2) + a_(3) +"........."+a_(n)),(a_(2))/(a_(1) + a_(3)+"..........."+a_(n)),"......",(a_(n))/(a_(1)+a_(2)+"........"a_(n-1)) are in :

If a_(1) , a_(2),"………",a_(n) are n non-zero real numbers such that ( a_(1)^(2) +a_(2)^(2) + "........."+a_(n-1)^(2) ) ( a_(2)^(2) + a_(3)^(2) + "........"+a_(n)^(2))le(a_(1) a_(2) + a_(2) a_(3) +".........." +a_(n-1) a_(n))^(2), a_(1), a_(2),".........",a_(n) are in :

If a_(1),a_(2),a_(3),a_(4),,……, a_(n-1),a_(n) " are distinct non-zero real numbers such that " (a_(1)^(2) + a_(2)^(2) + a_(3)^(2) + …..+ a_(n-1)^(2))x^2 + 2 (a_(1)a_(2) + a_(2)a_(3) + a_(3)a_(4) + ……+ a_(n-1) a_(n))x + (a_(2)^(2) +a_(3)^(2) + a_(4)^(2) +......+ a_(n)^(2)) le 0 " then " a_(1), a_(2), a_(3) ,....., a_(n-1), a_(n) are in

If a_(1),a_(2),a_(3),a_(4),,……, a_(n-1),a_(n) " are distinct non-zero real numbers such that " (a_(1)^(2) + a_(2)^(2) + a_(3)^(2) + …..+ a_(n-1)^(2))x^2 + 2 (a_(1)a_(2) + a_(2)a_(3) + a_(3)a_(4) + ……+ a_(n-1) a_(n))x + (a_(2)^(2) +a_(3)^(2) + a_(4)^(2) +......+ a_(n)^(2)) le 0 " then " a_(1), a_(2), a_(3) ,....., a_(n-1), a_(n) are in

A sequence of real numbers a_(1), a_(2), a_(3),......a_(n+1) is such that a_(1)=0, |a_(2)|=|a_(1)+1|, |a_(3)|=|a_(2)+1| Then (a_(1)+a_(2)+.........+a_(n))/(n) cannot be equal to.

If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .

If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .

If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .