Find no. of natural numbers which equal to the sum of squares of their

1.	Find no. of natural numbers which equal to the sum of squares of their digits
Answer» Answer: The family of natural numbers includes all the counting numbers, starting from 1 till infinity. If n consecutive natural numbers are 1, 2, 3, 4, …, n, then the sum of squared ‘n’ consecutive natural numbers is REPRESENTED by 12 + 22 + 32 + … + n2. In short, it is denoted by the notation Σn2. The formula for the ADDITION of squares of natural numbers is GIVEN below: Σn2 = [n(n+1)(2n+1)]/6 Step-by-step explanation: The family of natural numbers includes all the counting numbers, starting from 1 till infinity. If n consecutive natural numbers are 1, 2, 3, 4, …, n, then the sum of squared ‘n’ consecutive natural numbers is represented by 12 + 22 + 32 + … + n2. In short, it is denoted by the notation Σn2. The formula for the addition of squares of natural numbers is given below: Σn2 = [n(n+1)(2n+1)]/6 pls mark me as brainlist

Find no. of natural numbers which equal to the sum of squares of their digits

Answer»

Answer:

The family of natural numbers includes all the counting numbers, starting from 1 till infinity. If n consecutive natural numbers are 1, 2, 3, 4, …, n, then the sum of squared ‘n’ consecutive natural numbers is REPRESENTED by 12 + 22 + 32 + … + n2.

In short, it is denoted by the notation Σn2. The formula for the ADDITION of squares of natural numbers is GIVEN below:

Σn2 = [n(n+1)(2n+1)]/6

Step-by-step explanation:

In short, it is denoted by the notation Σn2. The formula for the addition of squares of natural numbers is given below:

Σn2 = [n(n+1)(2n+1)]/6

pls mark me as brainlist

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