Telescopic generations!

Let $\displaystyle S_m = \sum_{n=0}^m \dfrac 1{2^n+\sqrt{2^m}}$ and let us check the first few $S_m$ .

\(\begin{array} {} S_{\color{blue}0} = \dfrac 1{2^0+\sqrt{2^0}} = \dfrac 12 & = \dfrac {{\color{blue}0}+1}{\sqrt{2^{{\color{blue}0}+2}}} \\ S_{\color{blue}1} = \dfrac 1{1+\sqrt 2} + \dfrac 1{2+\sqrt 2} = \dfrac 1{\sqrt 2} & = \dfrac {{\color{blue}1}+1}{\sqrt{2^{{\color{blue}1}+2}}} \\ S_{\color{blue}2} = \dfrac 1{1+2} + \dfrac 1{2+2} + \dfrac 1{4+2} = \dfrac 34 & = \dfrac {{\color{blue}2}+1}{\sqrt{2^{{\color{blue}2}+2}}} \end{array} \)

It appears that we can claim that $S_{\color{#3D99F6}m} = \dfrac {{\color{#3D99F6}m}+1}{\sqrt{2^{{\color{#3D99F6}m}+2}}}$ . Let us prove by induction that the claim is true for all $m \ge 0$ .

Proof:

• We already know that the claim is true for $m=0$ and $m=1$ .

• Assuming that the claim is true for $m$ , then we have:

$\begin{aligned} \quad S_{\color{#3D99F6}m+2} & = \sum_{n=0}^{m+2} \frac 1{2^n+\sqrt{2^{m+2}}} \\ & = \frac 1{1+2^\frac {m+2}2} + \sum_{n={\color{#D61F06}1}}^{m+{\color{#D61F06}1}} \frac 1{2^n+2^\frac {m+2}2} + {\color{#3D99F6}\frac 1{2^{m+2}+2^\frac {m+2}2}} \\ & = \frac 1{2^\frac {m+2}2+1} + {\color{#3D99F6}\frac 1{2^\frac {m+2}2\left(2^\frac {m+2}2+1 \right)}} + \frac 12 \sum_{n=1}^{m+1} \frac 1{2^{n-1}+2^\frac m2} \\ & = \frac {2^\frac {m+2}2+1}{2^\frac {m+2}2\left(2^\frac {m+2}2+1 \right)} + \frac 12 \sum_{n={\color{#D61F06}0}}^{{\color{#D61F06}m}} \frac 1{2^n+2^\frac m2} \\ & = \frac 1{2^\frac {m+2}2} + \frac 12 S_m \\ & = \frac 1{\sqrt{2^{m+2}}} + \frac {m+1}{2 \sqrt{2^{m+2}}} \\ & = \frac {{\color{#3D99F6}m+2}+1}{ \sqrt{2^{{\color{#3D99F6}m+2}+2}}} \end{aligned}$

So the claim is also true for $m+2$ , since the claim is true for $m=0$ , it is also true for all even $m>0$ , likewise since it is true for $m=1$ , it is also true for all odd $m>1$ . Therefore, the claim is true for all $m \ge 0$ .

Therefore, $S_{1967} = \dfrac {1948}{\sqrt{2^{1949}}} = \dfrac {487}{\sqrt{2^{1945}}} \implies A+B = 487 + 1945 = \boxed{2432}$

$Let\quad a\quad be\quad \sqrt { { 2 }^{ 1947 } } \\ and\quad let\quad x\quad be\quad \sum _{ 0 }^{ 1947 }{ \frac { 1 }{ a\quad +\quad { 2 }^{ n } } } (\quad the\quad original\quad question)\\ if\quad we\quad observe\quad this\quad series\quad from\quad backwards\quad we\quad can\quad rewrite\quad it\quad as,\\ \sum _{ 0 }^{ 1947 }{ \frac { 1 }{ a(\frac { a }{ { 2 }^{ n } } +1) } } =\quad \frac { { 2 }^{ n } }{ a(a\quad +\quad { 2 }^{ n }) } =\quad \frac { 1 }{ a } -\quad \frac { 1 }{ a+{ 2 }^{ n } } =\quad x\\ \Rightarrow \quad \sum _{ 0 }^{ 1947 }{ \frac { 1 }{ a } } -\sum _{ 0 }^{ 1947 }{ \frac { 1 }{ a+{ 2 }^{ n } } } =\quad x\\ \Rightarrow \frac { 1948 }{ a } =\quad 2x\\ \Rightarrow \quad \frac { 974 }{ a } \quad =\quad x\\ putting\quad in\quad the\quad value\quad of\quad a\quad we\quad get\quad x\quad and\quad hence\quad the\quad required\quad answer.$