If S_n denotes the sum to n terms of the series 1<₌n<₌9

1+22+333+......+999+...... , then for n₌>2

(9 times)

a). S_n-S_n-1=1\9(10ⁿ-n²+n) b).S

c). 9(S_n-S_n-1)=n(10ⁿ-1) d).S₃= 356

Dear swapnil ,

there is a method of solving objective questions called solving by elimination. although we can try about solving for S_n -S_n-1 but in case of objective exams where speed matters it is better to use the former one.

so here , S_n -S_n-1 = a_n  , i.e. nth term.

a) apply n=2 , we get a₂ not equal to 22 so the option is eliminated.

b) not clearly mentioned by you.

c)apply two a2 is 22 but it you shud check also for n =9 where it gives wrong answer so option is eliminated.

d) manually see it is right .

so option d is correct.

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.


Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Win exciting gifts by answering the questions on Discussion Forum