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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


Answer: 4641.

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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Categories
Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


Answer: 399960.

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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College Mathematics

Sum Of Series: IIT JAM 2018 Problem 13

What about a small warm up MCQ!!!!!

[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
Topic
Series
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

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.

The will look more easy if we take a look into the knowledge graph

Let’s have a look into the graphs 

Do You Know ????

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