The Pivotals of Mathematics

Algebra Level 5

$a+b+c=\lambda$ for reals $a,b,c$ . We define $\displaystyle t_n= \sum \limits_{cyc} a^{2n+1}$ for $n\in \text{N}$ . Further it's given that $t_1={\lambda}^3$ . Find the value of $\lambda$ if $\displaystyle \sum_{n=1}^{\infty} t_n=\frac{1}{6}$ .

Note: This is my original problem belonging to the set Questions I've Made © . Give a try to its sister problem

The answer is 0.5.

1 solution

Joel Tan
Feb 23, 2015

$t_{1}=a^{3}+b^{3}+c^{3}=(a+b+c)^{3}$ .

Expanding and subtracting $a^{3}+b^{3}+c^{3}$ from both sides of the equation, then factorising gives the key result: $3 (a+b)(b+c)(c+a)=0$

Thus one of $(a+b), (b+c), (c+a) =0$ .

WLOG let $a+b=0$ then $t_{n}=a^{2n+1}+(-a)^{2n+1}+c^{2n+1}=c^{2n+1}$ .

Let the infinite sum be S. Summing all $t_{i}$ gives $\frac {c^{3}}{1-c^{2}}$ . This is by the formula for geometric progressions. Another condition for $c$ appears, which is that the absolute value of c <1. But solving the equation $S=\frac {1}{6}$ gives only one real solution $c=0.5 \implies a+b+c=0.5$ .

How can there be no loss in generality when assumed that $a+b = 0$ ? Please explain !

Venkata Karthik Bandaru - 6 years, 2 months ago

Because he's saying that the same logic can be applied for $b + c = 0$ and $a + c = 0$

Kyle Coughlin - 5 years, 11 months ago

When I read your solution my whole hard work on this question went in vain. This is not fair!

Aakash Khandelwal - 5 years, 10 months ago

In front of your solution mine is just madness of millennium . I expanded the expression

Aakash Khandelwal - 5 years, 10 months ago

The Pivotals of Mathematics

Note: This is my original problem belonging to the set Questions I've Made © . Give a try to its sister problem

The answer is 0.5.

1 solution

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