1q1
TAKE 2/9 COMMON SO U GET : 2/9(9+99+999+.....)
2/9(10+100+1000+...N TIMES.. -1-1-1-1-1.....N TIMES..)
NOW ITS AN INFINITE GP AND SIMPLE SUBTRCTN!!!!
HOPE U GOT IT!!!
1. If N= 2+22+222+.... upto 30000 terms
then find (a) last three digits of N
(b)last four digits of N
(c)last digit of N
2. (2)(2) + (4)(4) +(7)(8) +(11)(16) + (16)(32) + .......... Find the
sum to n terms of the series
3. Sum to (n+1) terms of the series
1+x/a1 + x(x+a1)/a1a2 + x(x+a1)(x+a2)/a1a2a3 + ......................
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7 Answers
1
12.) any kth term : t(k) = [2 + {(1+2 +3+....+k)-1} ] X [2X2k-1]
i thnk now u can solve it...
13.) t(k) = x/ak-1 (x/a1 -1)(x/a2 -1).....(x/ak-1 -1)
ne hint from here ?
13.) t(k) = x/ak-1 (x/a1 -1)(x/a2 -1).....(x/ak-1 -1)
ne hint from here ?
62Small correctino to madeforIIT's explanation.....
Q1
last 3 digits will be same as
last 3 digits of ... 2+22+222(N-2)
same for 4 digits :)
62S= (2)(2) + (4)(4) +(7)(8) +(11)(16) + (16)(32) +..........+
2S= 2.4 +4.8 + 7.16 +............................... + 2Tn
Take difference,
-S=4 +(2.4+3.8+4.16+...................) - 2Tn
Now find the sum in the bracket and Tn
I Hope u can finish this one now..
623. Sum to (n+1) terms of the series
1+x/a1 + x(x+a1)/a1a2 + x(x+a1)(x+a2)/a1a2a3 + ......................
this is sort of easy to be pulled off..
Solve by reccurrsion..
take 2 terms then add the 3rd ... then add the 4th.. .you will see how easy this one is :)
