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Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Sets and Integers.

For each positive integer n consider the set \(S_n\) defined as follows \(S_1\)={1}, \(S_2\)={2,3}, \(S_3\)={4,5,6}, ..., and , in general, \(S_{n+1}\) consists of n+1 consecutive integers the smallest of which is one more than the largest integer in \(S_{n}\). Then the sum of all the integers in \(S_{21}\) equals

- 1113
- 4641
- 53361
- 5082

Sets

Integers

Sum

But try the problem first...

Answer: 4641.

Source

Suggested Reading

B.Stat Objective Problem 121

Challenges and Thrills of Pre-College Mathematics by University Press

First hint

\(S_1\) has 1 element

\(S_2\) has 2 element

.....

\(S_{20}\) has 20 element

Second Hint

So number of numbers covered=1+2+3+...+20

sum =\(\frac{(20)(21)}{2}\)=210

\(S_{21}\) has 21 elements with first element= 211

Final Step

sum of n terms of a.p series with common difference d,

\(sum=\frac{n}{2}[2a+(n-1)d]\)

Then in our given question no of terms n=21 and c.d =1

Then the sum of elements=\(\frac{21}{2}[(2)(211)+(20)(1)]\)=4641.

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