Question

Calculati, va rog, urmatoarea suma:
S= 1+ 2·2^1 + 3·2^2 + 4·2^3 + ... + 2016·2^2015

Answer 1

$\boxed{\boxed{1\cdot 2^0+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+n\cdot2^{n-1} = 1+(n-1)\cdot 2^n}} \\ \\ \\ \Rightarrow 1+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+2016\cdot2^{2015} = \\ =1\cdot 2^0+2\cdot2^1+3\cdot2^2+4\cdot2^3 +...+ 2016\cdot2^{2016-1} = 1+(2016-1)\cdot 2^{2016}= \\ =1+2015\cdot 2^{2016} \\ \\ \Rightarrow \boxed{1+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+2016\cdot2^{2015} = 1+2015\cdot 2^{2016}}$


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