Sums (With Solutions)

These sums were taken from the book Number Theory by George E. Andrews

What is the equation for each of the following sums that gives the sum for a given $n$:

(1)  $1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$ since:

• $2(1 + 2 + 3 + \dots + n) =$ $([1+ 2 + 3 + \dots + n] + [n + (n-1) + (n-2) + \dots + 1]) =$ $(1 + n) + (2+[n-1]) + (3+[n-2]) + \dots + (n + 1) = (n)(n+1)$ 

(2)  $1^2 + 2^2 + 3^2 + \dots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ since:

• Binomial Theorem: $(x+y)^p = \sum\limits_{k=0}^{p}{p \choose k}x^{p-k}{y^{k}}$
• $(n-1)^3 = n^3 -3n^2 +3n - 1$   
• $n^3 - (n-1)^3 = 3n^2 - 3n + 1$ 
• $n(n+1)(2n+1) = (2n+1)(n^2 + n) = 2n^3 + 2n^2 + n^2 + n =$ $2n^3 + 3n^2 + n$
• $n^3 =$  $n^3 - (n-1)^3 + (n-1)^3 - (n-2)^3 + \dots + 1^3 - 0 =$  $3(n^2 + (n-1)^2 + \dots 1^2) - 3(n + (n-1) + \dots + 1) + n(1) =$  $3(1^2 + 2^2 + \dots + n^2) - 3(\dfrac{n(n+1)}{2}) + n$ 
• $n^3 + 3\left(\dfrac{n(n+1)}{2}\right) - n =$  $\dfrac{2n^3 + 3n^2 + 3n - 2n}{2} =$  $\dfrac{2n^3 + 3n^2 + n}{2} =$ $\dfrac{(n)(n+1)(2n+1)}{2} =$ $3(n^2 + (n-1)^2 + \dots 1^2)$  

(3)  $1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$ since:

• $(n-1)^4 = n^4 - 4n^3 + 6n^2 - 4n + 1$ 
• $n^4 - (n-1)^4 = 4n^3 - 6n^2 + 4n - 1$
• $[n(n+1)]^2 = n^2(n^2 + 2n + 1) = n^4 + 2n^3 + n^2$
• $n^4 = n^4 - (n-1)^4 + (n-1)^4 - (n-2)^4 + \dots + 1^4 - 0^4 =$ $4(n^3 + (n-1)^3 + \dots + 1^3) - 6(n^2 + (n-1)^2 + \dots + 1^2) + 4(n + (n-1) + \dots + 1) - n =$  $4(1^3 + 2^3 + \dots + n^3)  - 6\left(\dfrac{2n^3 + 3n^2 +n}{6}\right) + 4\left(\dfrac{n(n+1)}{2}\right) -n$
• $n^4 + 6\left(\dfrac{2n^3 + 3n^2 +n}{6}\right) - 4\left(\dfrac{n^2 + n)}{2}\right)  + n =$ $n^4 + 2n^3 + 3n^2 + n - 2n^2 -2n +n =$  $n^4 + 2n^3 + n^2 =$ $[n(n+1)]^2 = 4(1^3 + 2^3 + \dots + n^3)$

(4)  $1 + x + x^2 + x^3 + \dots + x^n = \dfrac{x^{n+1} - 1}{x-1}$ since:

• $x(1 + x + x^2 + x^3 + \dots + x^n) = x + x^2 + x^3 + x^4 + \dots + x^{n+1}$
• $(x-1)(1 + x + x^2 + x^3 + \dots + x^n) = x^{n+1} - 1$

(5) $1\times2 + 2\times3 + 3\times4 + \dots + (n)(n+1) = \dfrac{n(n+1)(n+2)}{3}$ since:

• $n(n+1) = n^2 + n$
• $1\times2 + 2\times3 + 3\times4 + \dots + (n)(n+1) =$ $(1^2 + 1) + (2^2 + 2) + (3^2 + 3) + \dots + + (n^2 + n) =$ $(1 + 2 + 3 + \dots + n) + (1^2 + 2^2 + 3^2 + \dots + n^2) =$ $\dfrac{n(n+1)}{2} + \dfrac{n(n+1)(2n+1)}{6} =$ $\dfrac{(3n + 2n^2 +n)(n+1)}{6} = \dfrac{(n^2+2n)(n+1)}{3}$

(6)  $1 + 3 + 5 + \dots + (2n-1) = n^2$ since:

• $1 + 3 + 5 + \dots + (2n-1) =$ $2(1 + 2 + 3  + \dots + n) - n =$ $n(n+1) - n = n^2 + n -n$  

(7)  $\dfrac{1}{1\times2} + \dfrac{1}{2\times3} + \dfrac{1}{3\times4} + \dots + \dfrac{1}{n(n+1)} = \dfrac{n}{n+1}$ since:

•  $\dfrac{1}{1\times2} + \dfrac{1}{2\times3} + \dfrac{1}{3\times4} + \dots + \dfrac{1}{n(n+1)} =$ $\dfrac{2-1}{1\times2} + \dfrac{3-2}{2\times3} + \dfrac{4-3}{3\times4} + \dots + \dfrac{(n+1)-n}{n(n+1)} =$ $\left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \dots + \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) =$ $1 - \dfrac{1}{n+1} = \dfrac{n+1 - 1}{n+1}$