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Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbers to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$

And then convert back into ordinary powers (by expansion, or using signed Stirling cycle numbers): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbers to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_1$$

And then convert back into ordinary powers (by expansion, or using signed Stirling cycle numbers): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling numbers of the second kind to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$

And then convert back into ordinary powers (by expansion, or using signed Stirling numbers of the first kind): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbers to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$

And then convert back into ordinary powers (by expansion, or using signed Stirling cycle numbers): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling numbers of the second kind to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$

And then convert back into ordinary powers (by expansion, or using signed Stirling numbers of the first kind): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling numbers of the second kind to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$

Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$

And then convert back into ordinary powers (by expansion, or using signed Stirling numbers of the first kind): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))$$

And then you can rearrange to get the answer you want.