Question

Put the following statements into order to prove the summation formula P(n) = (?_{i=0}^{n} i^2 = frac{n(n+1)(2n+1)}{6}) for all n ? ?. Put N next to the statements that should not be used. 1. By adding (n + 1)^2 to both sides of P(n), we get 2. Now suppose we have proved the statement P(n) for all n ? ?. 3. Now suppose we have proved the statement P(n) for some n ? ?. 4. The statement is true for n = 1 because in that case, both sides of the equation are 1. 5. ?_{i=0}^{n} i^2 + (n + 1)^2 = ?_{i=0}^{n+1} i^2 = frac{n(n+1)(2n+1)}{6} + (n + 1)^2 6. The statement is true for n = 1 because in that case, both sides of the equation are 0. 7. ?_{i=0}^{n} i^2 + n + 1 = ?_{i=0}^{n+1} i^2 = frac{n(n+1)(2n+1)}{6} + n + 1 8. By adding n + 1 to both sides of P(n), we get 9. Thus, we have proved the statement P(n + 1) = (?_{i=0}^{n+1} i^2 = frac{(n+1)(n+2)(2n+3)}{6}) 10. We simplify the right side: frac{n(n+1)(2n+1)}{6} + (n + 1)^2 = (n + 1)(frac{n(2n+1)}{6} + n + 1) = frac{(n+1)(n+2)(2n+3)}{6} 11. We simplify the right side: frac{n(n+1)(2n+1)}{6} + (n + 1) = (n + 1)(frac{n(2n+1)}{6} + 1) = frac{(n+1)(n+2)(2n+3)}{6}

          Put the following statements into order to prove the summation formula P(n) = (?_{i=0}^{n} i^2 = frac{n(n+1)(2n+1)}{6}) for all n ? ?. Put N next to the statements that should not be used.

1. By adding (n + 1)^2 to both sides of P(n), we get
2. Now suppose we have proved the statement P(n) for all n ? ?.
3. Now suppose we have proved the statement P(n) for some n ? ?.
4. The statement is true for n = 1 because in that case, both sides of the equation are 1.
5. ?_{i=0}^{n} i^2 + (n + 1)^2 = ?_{i=0}^{n+1} i^2 = frac{n(n+1)(2n+1)}{6} + (n + 1)^2
6. The statement is true for n = 1 because in that case, both sides of the equation are 0.
7. ?_{i=0}^{n} i^2 + n + 1 = ?_{i=0}^{n+1} i^2 = frac{n(n+1)(2n+1)}{6} + n + 1
8. By adding n + 1 to both sides of P(n), we get
9. Thus, we have proved the statement P(n + 1) = (?_{i=0}^{n+1} i^2 = frac{(n+1)(n+2)(2n+3)}{6})
10. We simplify the right side: frac{n(n+1)(2n+1)}{6} + (n + 1)^2 = (n + 1)(frac{n(2n+1)}{6} + n + 1) = frac{(n+1)(n+2)(2n+3)}{6}
11. We simplify the right side: frac{n(n+1)(2n+1)}{6} + (n + 1) = (n + 1)(frac{n(2n+1)}{6} + 1) = frac{(n+1)(n+2)(2n+3)}{6}

$Put the following statements into order to prove the summation formula P(n) = (?i=0^n i^2 = fracn(n+1)(2n+1)6) for all n ? ?. Put N next to the statements that should not be used. 1. By adding (n + 1)^2 to both sides of P(n), we get 2. Now suppose we have proved the statement P(n) for all n ? ?. 3. Now suppose we have proved the statement P(n) for some n ? ?. 4. The statement is true for n = 1 because in that case, both sides of the equation are 1. 5. ?i=0^n i^2 + (n + 1)^2 = ?i=0^n+1 i^2 = fracn(n+1)(2n+1)6 + (n + 1)^2 6. The statement is true for n = 1 because in that case, both sides of the equation are 0. 7. ?i=0^n i^2 + n + 1 = ?i=0^n+1 i^2 = fracn(n+1)(2n+1)6 + n + 1 8. By adding n + 1 to both sides of P(n), we get 9. Thus, we have proved the statement P(n + 1) = (?i=0^n+1 i^2 = frac(n+1)(n+2)(2n+3)6) 10. We simplify the right side: fracn(n+1)(2n+1)6 + (n + 1)^2 = (n + 1)(fracn(2n+1)6 + n + 1) = frac(n+1)(n+2)(2n+3)6 11. We simplify the right side: fracn(n+1)(2n+1)6 + (n + 1) = (n + 1)(fracn(2n+1)6 + 1) = frac(n+1)(n+2)(2n+3)6$

Added by Justin C.

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