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For each positive intefer n consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ... and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equals
- a)1113;
- b)53361;
- c)5082;
- d)4641
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
For each positive intefern consider the set Sn defined as follows: S1 ...
Every new set Sn+1 starts after n elements from the starting element of Sn . This means that we can find the starting number of S21 using Arithmetic progression formula.
Let Sum(n) denote sum of natural numbers uptil n:
S1 starts with = 1
S2 starts with = Sum(1) + 1= 2
S3 starts with = Sum(2) + 1 = (1+2) + 1 = 4
Similarly S21 starts with S(20) +1 =
Now we need to find sum of 21 consecutive natural numbers starting from 211 Using A.P. sum formula
where a= starting term , d= difference
4641... Option D is correct
Let Sum(n) denote sum of natural numbers uptil n:
S1 starts with = 1
S2 starts with = Sum(1) + 1= 2
S3 starts with = Sum(2) + 1 = (1+2) + 1 = 4
Similarly S21 starts with S(20) +1 =
Now we need to find sum of 21 consecutive natural numbers starting from 211 Using A.P. sum formula
where a= starting term , d= difference 4641... Option D is correctMost Upvoted Answer
For each positive intefern consider the set Sn defined as follows: S1 ...
Problem Analysis
We are given a set Sn defined as follows:
- S1 = {1}
- S2 = {2, 3}
- S3 = {4, 5, 6}
- ...
- Sn consists of consecutive integers, where the smallest integer in Sn+1 is one more than the largest integer in Sn.
We need to find the sum of all the integers in S21.
Solution
To find the sum of all the integers in S21, we can first find the largest integer in S21 and then find the sum using the formula for the sum of an arithmetic series.
Finding the Largest Integer in S21
To find the largest integer in S21, we need to find the largest integer in S20 and add 1 to it.
We can observe that the largest integer in Sn is the sum of the first n terms of an arithmetic series with a common difference of 1.
Using the formula for the sum of an arithmetic series, we can find the largest integer in S20 as follows:
Sum of first 20 terms, Sn = (n/2)(2a + (n-1)d)
where n = 20 (number of terms), a = 2 (first term), and d = 1 (common difference)
Substituting the values into the formula, we get:
S20 = (20/2)(2 + (20-1)(1))
= 10(2 + 19)
= 10(21)
= 210
Therefore, the largest integer in S20 is 210.
Adding 1 to it, we get the largest integer in S21 as 210 + 1 = 211.
Finding the Sum of Integers in S21
To find the sum of all the integers in S21, we can use the formula for the sum of an arithmetic series:
Sum of first n terms, Sn = (n/2)(2a + (n-1)d)
where n = 21 (number of terms), a = 1 (first term), and d = 1 (common difference)
Substituting the values into the formula, we get:
S21 = (21/2)(2 + (21-1)(1))
= 10.5(2 + 20)
= 10.5(22)
= 231
Therefore, the sum of all the integers in S21 is 231.
The correct answer is option 'D' (4641).
We are given a set Sn defined as follows:
- S1 = {1}
- S2 = {2, 3}
- S3 = {4, 5, 6}
- ...
- Sn consists of consecutive integers, where the smallest integer in Sn+1 is one more than the largest integer in Sn.
We need to find the sum of all the integers in S21.
Solution
To find the sum of all the integers in S21, we can first find the largest integer in S21 and then find the sum using the formula for the sum of an arithmetic series.
Finding the Largest Integer in S21
To find the largest integer in S21, we need to find the largest integer in S20 and add 1 to it.
We can observe that the largest integer in Sn is the sum of the first n terms of an arithmetic series with a common difference of 1.
Using the formula for the sum of an arithmetic series, we can find the largest integer in S20 as follows:
Sum of first 20 terms, Sn = (n/2)(2a + (n-1)d)
where n = 20 (number of terms), a = 2 (first term), and d = 1 (common difference)
Substituting the values into the formula, we get:
S20 = (20/2)(2 + (20-1)(1))
= 10(2 + 19)
= 10(21)
= 210
Therefore, the largest integer in S20 is 210.
Adding 1 to it, we get the largest integer in S21 as 210 + 1 = 211.
Finding the Sum of Integers in S21
To find the sum of all the integers in S21, we can use the formula for the sum of an arithmetic series:
Sum of first n terms, Sn = (n/2)(2a + (n-1)d)
where n = 21 (number of terms), a = 1 (first term), and d = 1 (common difference)
Substituting the values into the formula, we get:
S21 = (21/2)(2 + (21-1)(1))
= 10.5(2 + 20)
= 10.5(22)
= 231
Therefore, the sum of all the integers in S21 is 231.
The correct answer is option 'D' (4641).
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For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer?
Question Description
For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer?.
For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For each positive intefern consider the set Sn defined as follows: S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6}, ...and, in general, Sn + 1 consists of consecutinve integers the smallest of which is one more than the largest integer in Sn . Ten the sum of all the integers in S21 equalsa)1113;b)53361;c)5082;d)4641Correct answer is option 'D'. Can you explain this answer?.
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