Sum of squares of first square of n natural numbers formula
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sum of the squares of first n natural numbers =n(n+1)÷2
Pranjal01:
It is the formula for the sum of n natural numbers
The difference between two numbers is not common
n(n+1)(2n+1)/6 is the sum of squares of first n natural numbers
How do you get this???
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Answer:
Sum of squares of first n natural numbers is: n{n+1}{2n+1}/6
Step-by-step explanation:
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