mihir66 wrote:
1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?
A) 100.2^100
B) 99.2^100 + 1
C) 99.2^99 + 99
D) 2^100
A) 100.2^100
B) 99.2^100 + 1
C) 99.2^99 + 99
D) 2^100
.......................
30 seconds solution:
1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = An odd number, because 1+ even(everything is a multiple of 2).
so the Answer must be an odd number.
so A and D eliminated.
Now the last term of the series is 100.2^99 which is just 98 greater than option C) 99.2^99 + 99. (subtract and see).
but this is a series of 100 terms and even the 5th term itself is 80 . so its not possible for option (C), ultimately we have 1 option and its (B)