Formulas for Sums: $\sum _{ i=1 }^{ n }{ i },\ \sum _{ i=1 }^{ n }{ { i }^{ 2 } },\ \sum _{ i=1 }^{ n }{ { i }^{ 3 } }$

Suppose

$${ S }_{ n }=1+2+3+\cdots+n=\sum _{ i=1 }^{ n }{ i }.$$

To determine the formula ${ S }_{ n }$ can be done in several ways:

Method 1: Gauss Way

$$\begin{aligned} { S }_{ n }&=1+2+3+\cdots+n\\ { S }_{ n }&=n+(n-1)+(n-2)+\cdots+1\quad (+\\ \hline \\ 2{ S }_{ n }&=(n+1)+(n+1)+\cdots+(n+1)\\ &=n(n+1)\\ \\ \Rightarrow { S }_{ n }&=\frac { n(n+1) }{ 2 } \end{aligned}$$

Method 2: Telescoping Pattern

Observe that

$${ k }^{ 2 }-{ (k-1) }^{ 2 }=2k-1.$$

Then the value of $k$ is run from $1$ to $n$ to obtain the following telescoping:

$$\begin{aligned} { 1 }^{ 2 }-{ 0 }^{ 2 }&=2(1)-1\\ { 2 }^{ 2 }-{ 1 }^{ 2 }&=2(2)-1\\ { 3 }^{ 2 }-{ 2 }^{ 2 }&=2(3)-1\\ &\vdots \\ { n }^{ 2 }-{ (n-1) }^{ 2 }&=2n-1\qquad \qquad(+\\ \hline \\ { n }^{ 2 }-{ 0 }^{ 2 }&=2(1+2+3+\cdots+n)-n\\ { n }^{ 2 }+n&=2(1+2+3+\cdots+n)\\ \frac { { n }^{ 2 }+n }{ 2 } &=1+2+3+\cdots+n\\ \Rightarrow 1+2+3+\cdots+n &=\frac { n(n+1) }{ 2 }. \end{aligned}$$

Evaluate

$$\frac { 1+2+3+\cdots+2017 }{ 1009 }.$$

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We have

$$\begin{aligned} \frac { \sum _{ i=1 }^{ 2017 }{ i } }{ 1009 } &=\frac { 1+2+3+\cdots+2017 }{ 1009 } \\ \\ &=\frac {\ \ \frac { 2017(2017+1) }{ 2 }\ \ }{ 1009 } \\\\ &=\frac {\ \ \frac { 2017(2018) }{ 2 }\ \ }{ 1009 } \\ \\ &=\frac { 2017\times 1009 }{ 1009 } \\ \\ &=2017.\ _\square \end{aligned}$$

Evaluate

$$2 + 4 + 6 + \cdots + 2n.$$

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We have

$$\begin{aligned} \sum _{ i=1 }^{ n }{ 2i } &=2+4+6+\cdots+2n\\ &=2(1+2+3+\cdots+n)\\ &=2\left( \frac { n(n+1) }{ 2 } \right) \\ &=n(n+1).\ _\square \end{aligned}$$

Evaluate

$$1+3+5+\cdots+(2n-1).$$

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We have

$$\begin{aligned} \sum _{ i=1 }^{ n }{ (2n-1) } &=\sum _{ i=1 }^{ n }{ 2n } -\sum _{ i=1 }^{ n }{ 1 } \\ &=2\sum _{ i=1 }^{ n }{ n } -n\\ &=2\times \frac { n(n+1) }{ 2 } -n\\ &=n(n+1)-n\\ &=n(n+1-1)\\ &={ n }^{ 2 }.\ _\square \end{aligned}$$

Suppose we have the following sum:

$${ S }_{ n }={ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+\cdots+{ n }^{ 2 }=\sum _{ i=1 }^{ n }{ { i }^{ 2 } }.$$

In getting the sum ${ S }_{ n },$ we can travel with a telescoping pattern.

Observe algebraic form

$${ k }^{ 3 }-{ (k-1) }^{ 3 }={ 3k }^{ 2 }-3k+1.$$

By running the value of $k$ from $1$ to $n$, we obtain

$$\begin{aligned} 1^3-0^3&=3\big(1^2\big)-3(1)+1\\ 2^3-1^3&=3\big(2^2\big)-3(2)+1\\ 3^3-2^3&=3\big(3^2\big)-3(3)+1\\ &\vdots \\ n^3-(n-1)^3&=3\big(n^2\big)-3n+1 \qquad \large( + \\ \hline \\ n^3-0^3&=3\big(1^2+2^2+3^2+\cdots+n^2\big)-3(1+2+3+\cdots+n)+n\\ n^3&=3\sum _{i=1}^{n}{i^2}-3\sum _{i=1}^{n}{ i } +n \\ \sum _{ i=1 }^{ n }{ { i }^{ 2 }} &=\frac { 1 }{ 3 } \left[ { n }^{ 3 }+3\sum _{ i=1 }^{ n }{ i } -n \right] \\ &=\frac { 1 }{ 6 } \left[ 2{ n }^{ 3 }+3n^{ 2 }+3n-2n \right] \\ &=\frac { 1 }{ 6 } \left[ { 2n }^{ 3 }+{ 3n }^{ 2 }+n \right] \\ &=\frac { 1 }{ 6 } n\big({ 2n }^{ 2 }+3n+1\big)\\ &=\frac { 1 }{ 6 } n(n+1)(2n+1)\\ &=\frac { n(n+1)(2n+1) }{ 3! }. \end{aligned}$$

Evaluate

$$\frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+\cdots+{ 2017 }^{ 2 } }{ 1009\cdot 1345 }.$$

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We have

$$\begin{aligned} \frac { \sum _{ i=1 }^{ 2017 }{ { i }^{ 2 } } }{ 1009\cdot 1345 } &=\frac {\ \ \frac { 2017(2017+1)(2\cdot 2017+1) }{ 3! }\ \ }{ 1009\cdot 1345 } \\ \\ &=\frac {\ \ \frac { 2017\cdot 2018\cdot 4035 }{ 2\cdot 3 }\ \ }{ 1009\cdot 1345 } \\\\ &=\frac { 2017\cdot 1009\cdot 1345 }{ 1009\cdot 1345 } \\ \\ &=2017.\ _\square \end{aligned}$$

Evaluate

$${ 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+\cdots+{ (2n) }^{ 2 }.$$

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We have

$$\begin{aligned} \sum _{ i=1 }^{ n } (2i)^2 &=\sum _{ i=1 }^{ n }\big({ 2 }^{ 2 }{ i }^{ 2 }\big) \\ &=4\sum _{ i=1 }^{ n }{ { i }^{ 2 } } \\ &=4\cdot \frac { n(n+1)(2n+1) }{ 6 } \\ &=\frac { 2n(n+1)(2n+1) }{ 3 }.\ _\square \end{aligned}$$

Let

$${ S }_{ n }={ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+\cdots+{ n }^{ 3 }.$$

In getting the sum $S_n,$ we can travel with a telescoping pattern.

Observe algebraic form

$${ k }^{ 4 }-{ (k-1) }^{ 4 }=4{ k }^{ 3 }-6{ k }^{ 2 }+4k-1.$$

Then the value of $k$ is run from $1$ to $n$ to obtain the following telescoping:

$$\begin{aligned} { 1 }^{ 4 }-{ 0 }^{ 4 }&=4\big({ 1 }^{ 3 }\big)-6\big({ 1 }^{ 2 }\big)+4(1)-1\\ { 2 }^{ 4 }-{ 1 }^{ 4 }&=4\big({ 2 }^{ 3 }\big)-6\big(2^{ 2 }\big)+4(2)-1\\ { 3 }^{ 4 }-{ 2 }^{ 4 }&=4\big({ 3 }^{ 3 }\big)-{ 6\big(3 }^{ 2 }\big)+4(3)-1\\ &\vdots \\ { n }^{ 4 }-{ (n-1) }^{ 4 }&=4{ n }^{ 3 }-6{ n }^{ 2 }+4n-1\qquad\qquad \large( + \\ \hline \\ { n }^{ 4 }&=4\sum _{ i=1 }^{ n }{ { i }^{ 3 } } -6\sum _{ i=1 }^{ n }{ { i }^{ 2 } } +4\sum _{ i=1 }^{ n }{ i } -n\\ \sum _{ i=1 }^{ n }{ { i }^{ 3 } } &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+6\sum _{ i=1 }^{ n }{ { i }^{ 2 } } -4\sum _{ i=1 }^{ n }{ i } +n \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+6\cdot \frac { n(n+1)(2n+1) }{ 6 } -4\cdot \frac { n(n+1) }{ 2 } +n \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+n(n+1)(2n+1)-2n(n+1)+n \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+n(n+1)(2n+1)-n\left\{ 2(n+1)-1 \right\} \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+n(n+1)(2n+1)-n(2n+1) \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+n(2n+1)(n+1-1) \right] \\ &=\frac { 1 }{ 4 } \left[ { n }^{ 4 }+{ n }^{ 2 }(2n+1) \right] \\ &=\frac { 1 }{ 4 } { n }^{ 2 }\left[ { n }^{ 2 }+2n+1 \right] \\ &=\frac { 1 }{ 4 } { n }^{ 2 }{ \left[ n+1 \right] }^{ 2 }\\ &={ \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 }. \end{aligned}$$

Evaluate

$$\frac { \sqrt { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+\cdots+{ 2017 }^{ 3 } } }{ 1009 }.$$

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We have

$$\begin{aligned} \frac { \sqrt { \sum _{ i=1 }^{ 2017 }{ { i }^{ 3 } } } }{ 1009 } &=\frac { \sqrt { { \left( \frac { 2017(2017+1) }{ 2 } \right) }^{ 2 } } }{ 1009 } \\ \\ &=\frac {\ \frac { 2017\cdot 2018 }{ 2 }\ }{ 1009 } \\\\ &=\frac { 2017\cdot 1009 }{ 1009 } \\ \\ &=2017.\ _\square \end{aligned}$$