IIT JEE , AIEEE Forum - Mathematics , Algebra

ayshwarya

i think dat dis is something wrong

given in induction f(k+1)=f(k) +(33.....3 )(k+1)times =f(k)+1/3 (99.....9)(k+1)times

dis underlined is my mistake

ie

how cud 33..3(k+1)times = 1/3(99...9) k+1 times

plz reply

ultimator

It is (1/3)*(99....9)(k+1 times) and not 1/(3*(99...99)(k+1)) times as u think.

RyuAmakusa

yes this q i have done there is no error in that initallu there are k+1 three's then there are k+1 nine's

ayshwarya

no its not dat my dout is in d dr it must b 3^(k+1) but how cud it b only 3

RyuAmakusa

no it is not 9*9*9..... it is something like 9lakh 99thousand 9hundreden.....it is something like that it is a single digit.

ayshwarya

thanx ruam