Getting back to 1947

Taking first and last terms,

$\frac { 1 }{ 1+{ 2 }^{ \sqrt { 1947 } } } +\frac { 1 }{ { 2 }^{ 1947 }+{ 2 }^{ \sqrt { 1947 } } }$

$=\frac { 1 }{ 1+{ 2 }^{ \sqrt { 1947 } } } +\frac { 1 }{ 1+{ 2 }^{ \sqrt { 1947 } } } \left( \frac { 1 }{ { 2 }^{ \sqrt { 1947 } } } \right)$

$=\left( 1+\frac { 1 }{ { 2 }^{ \sqrt { 1947 } } } \right) \left( \frac { 1 }{ 1+{ 2 }^{ \sqrt { 1947 } } } \right)$

$=\left( \frac { { 2 }^{ \sqrt { 1947 } }+1 }{ { 2 }^{ \sqrt { 1947 } } } \right) \left( \frac { 1 }{ 1+{ 2 }^{ \sqrt { 1947 } } } \right)$

$=\frac { 1 }{ { 2 }^{ \sqrt { 1947 } } }$

Similarly pairing remaining terms (Note that there is no term without a pair), we get,

$\sum _{ n=0 }^{ 1947 }{ \frac { 1 }{ { 2 }^{ n }+\sqrt { { 2 }^{ 1947 } } } }$

= $\frac { 974 }{ { 2 }^{ \sqrt { 1947 } } }$

= $\frac { 487 \sqrt{2}}{ { 2 }^{ { 973 } } }$

Hence, $A + B = 1460$ .

Check out this

You want this:

$\sum _{n=0}^{1947} \frac{1}{2^n+\sqrt{2^{1947}}}=\frac{974 \sqrt{2}}{2\times\ 2^{973}}$