Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Sets and Integers.
Sets and Integers ( B.Stat Objective Question )
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
Key Concepts
Sets
Integers
Sum
Check the Answer
Answer: 4641.
B.Stat Objective Problem 121
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
\(S_1\) has 1 element
\(S_2\) has 2 element
.....
\(S_{20}\) has 20 element
So number of numbers covered=1+2+3+...+20
sum =\(\frac{(20)(21)}{2}\)=210
\(S_{21}\) has 21 elements with first element= 211
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.
Let $a_{n}=n+\frac{1}{n}, n \in \mathbb{N} .$ Then the sum of the series $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{a_{n+1}}{n !}$ is -
(A) $e^{-1}-1$
(B) $e^{-1}$
(C) $1-e^{-1}$
(D) $1+e^{-1}$
Start with hints
Do you really need a hint? Try it first!
[Hint 1] Consider $a_n = n + \frac{1}{n} , n \in \mathbb{N} $
We have to use $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!}+.....$
Specifically $e^{1} = 1+ \frac{1}{1!} + \frac{1}{2!}+....$
And $ e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + .......$
Do you want to play with it.
[Hint 2]$ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1 + \frac{1}{n+1}}{n!} $ = $ \sum_{n=1}^{\infty}[ (-1)^{n+1} \frac{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]$
Now we will be breaking it term by term for the ease of calculation. Can you do it from here?
[Hint 3]
$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} $
= $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$
= $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1] $ = $e^{-1} + 1 $ So option (D) is our required answer.
Let's have a look into the graphs
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