4 (a) By considering $(2 r+1)^{3}-(2 r-1)^{3}$, use the method of differences to prove that

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the expression $(2r + 1)^3 - (2r - 1)^3$. By expanding both cubes, we can simplify this expression. ‖ ‖ step 2 ⋮ The expansion gives us $8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2$. ‖ ‖ step 3 ⋮ Now, we can express the sum of $r^2$ from $r=1$ to $n$ using the formula: $\sum_{r=1}^{n} r^2 = \frac{1}{6} n (n + 1)(2n + 1)$. ‖ ‖ step 4 ⋮ This formula can be derived using the method of differences, which involves finding the difference between consecutive terms and summing them up. ‖ ∻Answer∻ ⚹ The sum of $r^2$ from $r=1$ to $n$ is $\frac{1}{6} n (n + 1)(2n + 1)$. ⚹ ∻Key Concept∻ ⚹ The method of differences is a technique used to derive summation formulas by analyzing the differences between terms. ⚹ ∻Explanation∻ ⚹ The answer shows that the sum of squares can be expressed in a closed form, which is useful for calculating sums efficiently. ⚹◊How can you express the sum of the squares of the first $ n $ integers using a formula, specifically in the form of $ \sum_{r=1}^{n} r^2 $?⍭ Generate me a similar question◊

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