Summation+nested radical

Calculus Level 3

If n = 1 n 2 2 n = S \displaystyle \sum_{n=1}^\infty \frac{ n^2 } { 2^n } = S ,
and f ( x ) = x + x + x + x + f(x) = \sqrt{ x + \sqrt{ x + \sqrt{ x + \sqrt{ x + \ldots}}}} ,
find f ( S ) f(S) .


The answer is 3.

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2 solutions

S = n = 1 n 2 2 n = n = 0 n 2 2 n S = n = 1 n 2 2 n = n = 0 ( n + 1 ) 2 2 n + 1 S = 2 S S = n = 0 ( n + 1 ) 2 2 n n = 0 n 2 2 n = n = 0 2 n + 1 2 n = n = 0 2 n 2 n + n = 0 1 2 n = 2 n = 0 2 n 2 n n = 0 2 n 2 n + n = 0 1 2 n = 2 n = 0 2 ( n + 1 ) 2 n + 1 n = 0 2 n 2 n + n = 0 1 2 n = n = 0 2 2 n + n = 0 1 2 n = 3 n = 0 1 2 n = 3 ( 1 1 1 2 ) S = 6 \begin{aligned} S & = \sum_{n=1}^\infty {\frac{n^2}{2^n}} = \sum_{n=0}^\infty {\frac{n^2}{2^n}} \\ S & = \sum_{n=1}^\infty {\frac{n^2}{2^n}} = \sum_{n=0}^\infty {\frac{(n+1)^2}{2^{n+1}}} \\ S & = 2S - S \\ & = \sum_{n=0}^\infty {\frac{(n+1)^2}{2^{n}}} - \sum_{n=0}^\infty {\frac{n^2}{2^n}} \\ & = \sum_{n=0}^\infty {\frac{2n+1}{2^n}} \\ & = \sum_{n=0}^\infty {\frac{2n}{2^n}} + \sum_{n=0}^\infty {\frac{1}{2^n}} \\ & = 2\sum_{n=0}^\infty {\frac{2n}{2^n}} - \sum_{n=0}^\infty {\frac{2n}{2^n}} + \sum_{n=0}^\infty {\frac{1}{2^n}} \\ & = 2\sum_{n=0}^\infty {\frac{2(n+1)}{2^{n+1}}} - \sum_{n=0}^\infty {\frac{2n}{2^n}} + \sum_{n=0}^\infty {\frac{1}{2^n}} \\ & = \sum_{n=0}^\infty {\frac{2}{2^n}} + \sum_{n=0}^\infty {\frac{1}{2^n}} = 3\sum_{n=0}^\infty {\frac{1}{2^n}} = 3\left(\frac{1}{1-\frac{1}{2}} \right) \\ \Rightarrow S & = 6 \end{aligned}

f ( S ) = 6 + 6 + 6 + 6 + . . . ( f ( S ) ) 2 = 6 + f ( S ) ( f ( S ) ) 2 f ( S ) 6 = 0 ( f ( S ) 3 ) ( f ( S ) + 2 ) = 0 Since f ( S ) > 0 f ( S ) = 3 \begin{aligned} f(S) & = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}} \\ \Rightarrow \left( f(S) \right)^2 & = 6 + f(S) \\ \left( f(S) \right)^2 - f(S) - 6 & = 0 \\ (f(S)-3)(f(S)+2) & = 0 \quad \text{Since } f(S) > 0 \\ \Rightarrow f(S) & = \boxed{3} \end{aligned}

Moderator note:

Simple standard solution.

Dear Chew-Seong, I did a similar method here.

Noel Lo - 6 years ago
Alex Delhumeau
Jun 3, 2015

S = n = 1 n 2 2 n S = n = 1 ( ( 2 n 1 ) m = n 1 2 m ) S = n = 1 2 n 1 2 n 1 S = 2 n = 1 2 n 1 2 n S = 2 ( 1 + 2 n = 2 m = n 1 2 m ) S = 2 ( 1 + 2 n = 2 1 2 n 1 ) S = 2 ( 1 + 2 1 ) S = 6 \begin{aligned} S & = \sum_{n=1}^\infty {\frac{n^2}{2^n}} \\ S & =\sum_{n=1}^\infty {((2n-1)\cdot\sum_{m=n}^\infty {\frac{1}{2^m}})} \\ S & = \sum_{n=1}^\infty {\frac{2n-1}{2^{n-1}}} \\ S & = 2\sum_{n=1}^\infty {\frac{2n-1}{2^n}} \\ S & = 2(1+2\sum_{n=2}^\infty {\sum_{m=n}^\infty {\frac{1}{2^m}}}) \\ S & = 2(1+2\sum_{n=2}^\infty {\frac{1}{2^{n-1}}}) \\ S & = 2\cdot(1+2\cdot1) \\ S & = 6 \end{aligned}

6 + 6 + 6 + 6 + = x , \Rightarrow \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}}=x, 6 + x = x x 2 x 6 = 0 x = 3 \therefore \sqrt{6+x}=x \Rightarrow x^2-x-6=0 \Rightarrow x=\boxed{3} .

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