A calculus problem by Prakul Sharma

Calculus Level 1

6 + 6 + 6 + 6 + ⋯ = ? \large \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \cdots}}}} = \, ?

6 6 ∞ \infty 3 3 − 2 -2

View wiki
Prakul Sharma by Prakul Sharma , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Prakul Sharma
Apr 16, 2016

Let √(6+.. = y we can write as √(6+y) = y squaring both sides, 6+y = y^2 solving the quadratic equation, we get 2 roots --> 3, -2 but since sum of positive numbers cannot be negative, therefore answer is 3

2 Helpful 0 Interesting 1 Brilliant 0 Confused

Simplicity is the best

6 + 6 + 6 + 6 + 6 + . . . ⇒ C = x + 6 ⇒ x 2 − x − 6 = 0 ⇒ ( x − 1 2 ) 2 = 25 4 ⇒ x = 3 □ \begin{aligned} \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{...}}}}}} &\Rightarrow C = \sqrt{x+6} \\&\Rightarrow x^2-x-6 = 0 \\&\Rightarrow \left(x-\frac{1}{2}\right)^2 = \frac{25}{4} \\&\Rightarrow x=3 \space \square \end{aligned}

ADIOS!!! \LARGE \text{ADIOS!!!}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...