
#1
January 10th, 2020,
03:42
Hello all,
In the attached picture there is an equation. I need to fill the general expression on the left hand side, and to prove by induction that the sum is equal to the expression in the right hand side.
I am not sure how to find the general expression. Can you kindly assist ?
Thank you !

January 10th, 2020 03:42
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MHB Journeyman
#2
January 10th, 2020,
09:47
$\dfrac{(1)^{n+1}(2n+1)}{n^2+n}$

#3
January 11th, 2020,
13:41
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 $ \displaystyle 0+ 2n+ (2/2)n(n1)= 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 $ \displaystyle (1)^{n+1}$ or $ \displaystyle (1)^{n1}$.

MHB Journeyman
#4
January 11th, 2020,
16:18
I saw sequence of denominators, $2 ,6,12,20,30,...$, as
$(1\cdot 2), (2 \cdot 3), (3 \cdot 4), ( 4 \cdot 5),(5 \cdot 6), ... , [n \cdot (n+1)] , ...$

#5
January 13th, 2020,
16:34
Thread Author
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...
Last edited by Yankel; January 13th, 2020 at 17:19.

MHB Journeyman
#6
January 13th, 2020,
18:41
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}$