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sum_(k=3)^(n)(k^(2)-3)

1717. ∑k=3n(k2−3) \sum_{k=3}^{n}\left(k^{2}-3\right)

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Full solution

Q. 1717. ∑k=3n(k2−3) \sum_{k=3}^{n}\left(k^{2}-3\right)
  1. Simplify and Separate Terms: Simplify the series expression to separate k2k^2 and −3-3: ∑k=3n(k2)−∑k=3n(3)\sum_{k=3}^{n}(k^2) - \sum_{k=3}^{n}(3).
  2. Evaluate First Term: Evaluate the first term ∑k=3n(k2)\sum_{k=3}^{n}(k^2) using the summation formula for k2k^2: ∑k=1n(k2)−∑k=12(k2)\sum_{k=1}^{n}(k^2) - \sum_{k=1}^{2}(k^2).
  3. Apply Sum of Squares Formula: Apply the formula for the sum of squares: ∑k=1n(k2)=n(n+1)(2n+1)6\sum_{k=1}^{n}(k^2) = \frac{n(n+1)(2n+1)}{6}.
  4. Calculate First 22 Terms: Calculate the sum of squares for the first 22 terms: ∑k=12(k2)=2(2+1)(2⋅2+1)6=2⋅3⋅56=5\sum_{k=1}^{2}(k^2) = \frac{2(2+1)(2\cdot2+1)}{6} = \frac{2\cdot3\cdot5}{6} = 5.
  5. Subtract First 22 Squares: Subtract the sum of the first 22 squares from the total sum of squares: n(n+1)(2n+1)6−5\frac{n(n+1)(2n+1)}{6} - 5.
  6. Evaluate Second Term: Evaluate the second term ∑k=3n(3)\sum_{k=3}^{n}(3) as 33 times the number of terms, which is (n−2)(n-2).
  7. Calculate Second Term: Calculate the second term: 3(n−2)=3n−63(n-2) = 3n - 6.
  8. Subtract to Get Final Sum: Subtract the second term from the first to get the final sum: (n(n+1)(2n+1))/6−5−(3n−6)(n(n+1)(2n+1))/6 - 5 - (3n - 6).
  9. Combine Terms to Simplify: Combine the terms to simplify the expression: (n(n+1)(2n+1))/6−3n+1(n(n+1)(2n+1))/6 - 3n + 1.

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