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sum_(j=1)^(2)(j^(2)+1)=

j=12(j2+1)= \sum_{j=1}^{2}\left(j^{2}+1\right)=

Full solution

Q. j=12(j2+1)= \sum_{j=1}^{2}\left(j^{2}+1\right)=
  1. Problem understanding and setup: Understand the problem and set up the summation.\newlineWe need to calculate the sum of the expression j2+1j^2 + 1 for each integer value of jj from 11 to 22.
  2. Calculation for j=1j=1: Calculate the sum for j=1j=1. When j=1j=1, the expression becomes 12+1=1+1=21^2 + 1 = 1 + 1 = 2.
  3. Calculation for j=2j=2: Calculate the sum for j=2j=2.\newlineWhen j=2j=2, the expression becomes 22+1=4+1=52^2 + 1 = 4 + 1 = 5.
  4. Total sum calculation: Add the results from Step 22 and Step 33 to find the total sum.\newlineTotal sum = 22 (from j=1j=1) + 55 (from j=2j=2) = 77.

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