Math, asked by agrawal, 1 year ago

if sn denotes the nth term of the series 2+3+6+11+18+...... then find t50.

Answers

Answered by zxc7
2
the diff of the terms of the series are in AP and the series we get from their diff is 1,3,5...

now observe that
2
2+1=3
2+1+3=6
2+1+3+5=11
.
.
.
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)

zxc7: sorry but can you please rephrase your question in a more accurate manner
agrawal: i m saying that u are saying that to get the answer we first find the 49th term of the series 1,3,5,....and then add 2 to it isnt it ?????
zxc7: Precisely
agrawal: so i m asking that why here we are finding the 49th term and not 50th term of 1,3,5,...
agrawal: and then add 2 to it
agrawal: please answer
zxc7: observe that the n-th term of the series 2+3+6+11 is equal to the sum upto the (n-1)th term of the series 1+3+5+7 plus 2 and hence we find the 49th term.for example the third term of the original series 6 is equal to the sum upto the (3-1)th term of the series 1+3+5+7 plus 2
agrawal: ya now i got it thank you very miuch
zxc7: np
Svaishnavsv66: Helllllllooooooooooo everyone................
Answered by AbdulHafeezAhmed
3

the diff of the terms of the series are in AP and the series we get from their diff is 1,3,5...

now observe that

2

2+1=3

2+1+3=6

2+1+3+5=11

.

.

.

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

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