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Sets and Integers | TOMATO B.Stat Objective 121

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

B.Stat Objective Problem 121

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

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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Problem on Digits | TOMATO B.Stat Objective 111

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Problem on Digits.

Problem on Digits ( B.Stat Objective Question )

The sum of all the distinct four digit numbers that can be formed using the digits 1, 2, 3, 4, 5 each digit appearing at most once is

• 399900
• 399960
• 0
• none of these

Key Concepts

Digits

Sum

Non-repeated

B.Stat Objective Problem 111

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

First hint

here every strings are 1234, 2345, 1345, 1235, 2345

Second Hint

for 1234, sum =66660

for 1235, sum=66+660+6600+66000=73326

for 2345, sum=84+840+8400+84000=93324

for 1345, sum=78+780+7800+78000=86658

for 1245, sum=72+720+7200+72000=79992

Final Step

then 66660+93324+73326+79992+86658=399960.

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[h5p id="14"]

Understand the problem

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) $1+e^{-1}$
Source of the problem
IIT JAM 2018 Problem 13
Series
Easy
Suggested Book
Real Analysis By S.K Mapa

Do you really need a hint? Try it first!

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
$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1 + \frac{1}{n+1}}{n!}$ = $\sum_{n=1}^{\infty}[ (-1)^{n+1} \frac{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]$   Now we will be breaking it term by term for the ease of calculation. Can you do it from here?
$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!}$ = $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$ = $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1]$ = $e^{-1} + 1$ So option (D) is our required answer.

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