Albert said:$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$
Happy new year to all MHB folkskaliprasad said:Albert and others Happy new year
we have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\dfrac{1}{a+b+c}$
$=>\frac{1}{a}+\frac{1}{b} = \frac{1}{a+b+c}- \frac{1}{c}$
$=>\frac{a+b}{ab} = - \frac{a+b}{c(a+b+c)}$
$=>(a+b) = 0\cdots(1)$ or $\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$=>c(a+b+c) + ab = 0$
or $(a+c)(b+c) = 0$
so from (1) and above we have $a= -b$ or $b= -c$ or $c= - a$
beacuse of symmetry taking $a = -b$ we have a + b+ c = c and both LHS and RHS of given expression $\frac{1}{c^{2n+1}}$ and same
hence proved
This equation is used to prove that the sum of the reciprocals of three numbers raised to the power of (2n+1) is equal to the reciprocal of the sum of those three numbers raised to the same power.
The value of n can be any positive integer or zero. This equation holds true for all values of n.
This equation can be proven using mathematical induction, where the base case is n = 0 and the inductive step is to assume the equation holds true for n = k and then prove it for n = k+1.
Yes, this equation can be used with any three numbers as long as they are non-zero and finite. It is also applicable for both positive and negative numbers.
This equation is significant in mathematics as it demonstrates the principle of mathematical induction and also shows the relationship between the sum of the reciprocals of numbers and the reciprocal of the sum of those numbers.