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## Answers

now observe that

2

2+1=3

2+1+3=6

2+1+3+5=11

.

.

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so the n-th term of the original series is basically sum of n-1 terms of the series (1+3+5...) +2

hence calculate sum of 49 terms of 1+3+5... which is an AP with a common diff of 2 which can easily be obtained by the formula (n/2(2a+(n-1)d)) and then add 2 to get the sum upto the 50th term=2403(calc error possible)