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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.

Categories

[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?
$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!}$ = $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$ = $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1]$ = $e^{-1} + 1$Â So option (D) is our required answer.

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#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuingÂ who wish to rediscover the world of mathematics.

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