According to arithmetic progression, natural numbers can be written down as 1, 2, 3, 4, 5, 6, 7, and 8 to 100. Basically, the sum of the first 100 natural numbers is equal to 5050.
How to Find the Sum of Natural Numbers 1 to 100? The sum of all natural numbers from 1 to 100 is 5050.
Therefore, the sum 1 + 2 + 3 + 4 + 5 + . . . . . . + 100 = 5050 .
The sum of the first 1000 positive integers is 500500.
Basically, the sum of the first 100 natural numbers is equal to 5050.
Thus, the sum of the first 1000 positive integers is 500500.
Hence the total number lying between 100 and 1000 which can be formed with digits 0, 1, 2, 3, 4, 5, 6, 7 are 294 numbers i.e. \[{}^{8}{{P}_{3}}-{}^{7}{{P}_{2}}\]. \[\therefore \] Option (d) is the correct answer. Note: In a question like this be careful as to subtract 3 digit numbers starting with zero.
For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12.
or, S=946.
Addition Property
If any number is added to infinity, the sum is also equal to infinity.
This is an arithmetic sequence since there is a common difference between each term. In this case, adding 1 to the previous term in the sequence gives the next term. In other words, an=a1+d(n−1) a n = a 1 + d ( n - 1 ) . This is the formula of an arithmetic sequence.
Solution Approach
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10... For here, we can conclude that the sum can be formulated as, Sum = +(n+1)/2 if n is odd. Sum = -(n)/2 if n is even.
Statement 1: the sum of first 30 terms of the sequence 1,2,4,7,11,16,22.... is 4520.
We have found the sum of n terms of the series 1, 3, 5, 7, …… is ${{n}^{2}}$. $\therefore$ The sum of series 1, 3, 5, 7, 9,…… up to n terms is ${{n}^{2}}$. Note: We can also use the sum of 'n' natural numbers to find the sum of the given series after calculating the ${{n}^{th}}$ term.
Answer: The sum of infinite GP series 1/2 , 1/4 , 1/8 , 1/16 ... is 1.
A set that has a finite number of elements is said to be a finite set, for example, set D = {1, 2, 3, 4, 5, 6} is a finite set with 6 elements.
Boole came to the rescue by ingeniously recognising that binary logical operations behaved in a way that's strikingly similar to our normal arithmetic operations, with a few twists. 1 + 1 = 1 (since "true OR true" is true). 1 x 1 = 1 (since "true AND true" is true). A x NOT A = 0 (since "true AND false" is false).
By the fundamental principle of counting, the required number of numbers lying between 100 & 1000 formed out of {0,1,2,3,4,5}=5×5×4=100.
Without repetition counting the number of 3digit numbers using digits from {0,1,2,3,4,5}: Count the number of 3-digit strings whose last digit is even. This gives 60−8=52 3-digit even numbers using digits from {0,1,2,3,4,5} without repetition. Save this answer.
∴ 16 numbers will be there.
Hence, the answer to this question is 20100. Note – In this problem you need to consider the series from 1 to 200 as an AP and use the formula of sum of 200 terms of an AP. Doing this and solving algebraically you will get the right answer.
Answer and Explanation: We get that the sum of the integers from 1 to 1000 is 500,500.
Answer: The sum of the series 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 is equal to 45.