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sum_(j=0)^(2)(j^(2))=

j=02(j2)= \sum_{j=0}^{2}\left(j^{2}\right)=

Full solution

Q. j=02(j2)= \sum_{j=0}^{2}\left(j^{2}\right)=
  1. Problem Statement: The problem asks for the sum of the squares of the integers from 00 to 22. This can be written as the sum of j2j^2 where jj ranges from 00 to 22.
  2. Calculate Square of First Integer: First, calculate the square of the first integer, which is 00. The square of 00 is 02=00^2 = 0.
  3. Calculate Square of Second Integer: Next, calculate the square of the second integer, which is 11. The square of 11 is 12=11^2 = 1.
  4. Calculate Square of Third Integer: Finally, calculate the square of the third integer, which is 22. The square of 22 is 22=42^2 = 4.
  5. Sum of Squares: Now, sum the squares of these integers: 0+1+40 + 1 + 4.
  6. Final Result: The sum is 0+1+4=50 + 1 + 4 = 5.

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