Question

if 100 more than the sum of n consecutive natural numbers is equal to the sum of the next n consecutive natural numbers find n​

Answer 1

Answer: This may help you

Step-by-step explanation:

Sum of natural, odd & even numbers

Sum of “n” natural numbers = \frac{n(n+1)}{2}

Sum of “n” natural even numbers = (n ) (n + 1)

Sum of “n” natural odd numbers = n 2

Sum of square of natural, odd & even numbers

Sum of square of “n” first or consecutive square natural numbers:

Sum of square of "n" first or consecutive square natural numbers | Sum of n Consecutive numbers

Sum of square of “n” first or consecutive odd numbers:

Sum of square of "n" first or consecutive odd numbers | Sum of n Consecutive numbersSum of square of “n” first or consecutive square even numbers:

Sum of square of "n" first or consecutive square even number | Sum of n Consecutive numbers

Sum of cube natural, odd & even numbers

Sum of cube of first or consecutive ” n” natural numbers:

Sum of square of "n" first or consecutive square even number | Sum of n Consecutive numbers

Sum of cube of first or consecutive ” n” even natural numbers = 2n2 (n + 1)2

Sum of cube of first or consecutive ” n” odd natural numbers = n2 (2n2 – 1)

The easy-to-use formula is

(n / 2)(first number + last number) = sum, where n is the number of integers.

Let's use the example of adding the numbers 1-100 to see how the formula works.

Find the sum of the consecutive numbers 1-100:

(100 / 2)(1 + 100)

50(101) = 5,050

Answer 2

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Since each integer in second sum is n more than corresponding integer in first sum and since there are n integers in each sum,sum_1+100=Sum_1+n^2=>n^2=100

So,n=10

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HOPE IT HELPS!!