# Proof with Induction 3/2-5/6+7/12-9/20+11/30-...

#### skeeter

##### Well-known member
MHB Math Helper
$\dfrac{(-1)^{n+1}(2n+1)}{n^2+n}$

#### HallsofIvy

##### Well-known member
MHB Math Helper
The numerators are clearly the odd numbers so: 2n+ 1.

The denominators are a little harder! I would have used "Newton's "divided difference" formula: adding a first term of "0", the "first differences" are 2- 0= 2, 6- 2= 4, 12- 6= 6, 20- 12= 8, 30- 20= 10; the "second differences" are 4- 2= 2, 6- 4= 2, 8- 6= 2, 10- 8= 2. Those are all "2" so all further "differences" are 0. The denominators are given by the quadratic $$0+ 2n+ (2/2)n(n-1)= n^2+ n$$.

Of course, since the +/- sign alternates we need -1 to a power. The first term, with n= 1, is positive so that can be either $$(-1)^{n+1}$$ or $$(-1)^{n-1}$$.

#### skeeter

##### Well-known member
MHB Math Helper
I saw sequence of denominators, $2 ,6,12,20,30,...$, as

$(1\cdot 2), (2 \cdot 3), (3 \cdot 4), ( 4 \cdot 5),(5 \cdot 6), ... , [n \cdot (n+1)] , ...$

#### Yankel

##### Active member
I tried proving this by induction using the general statement that skeeter wrote, but I couldn't do it.

I am stuck at the n=k+1 stage...

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#### skeeter

##### Well-known member
MHB Math Helper
note $1 + \dfrac{(-1)^{n+1}}{n+1} = \dfrac{(n+1) + (-1)^{n+1}}{n+1}$

${\color{red}{\dfrac{3}{2} - \dfrac{5}{6} + \dfrac{7}{12} - \dfrac{9}{20} + ... + \dfrac{(-1)^{n+1}(2n+1)}{n(n+1)}}} + \dfrac{(-1)^{(n+1)+1}[2(n+1)+1]}{(n+1)[(n+1)+1]}$

${\color{red}\dfrac{(n+1) + (-1)^{n+1}}{n+1}} + \dfrac{(-1)^{n+2}(2n+3)}{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}(n+2)}{(n+1)(n+2)} + \dfrac{(-1)^{n+2}(2n+3)}{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}(n+2) - (-1)^{n+1}(2n+3) }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}[(n+2) - (2n+3)] }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+2}(n+1) }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2)}{(n+1)(n+2)}+ \dfrac{(-1)^{n+2}(n+1)}{(n+1)(n+2)}$

$1 + \dfrac{(-1)^{(n+1)+1}}{(n+1)+1}$