What is the last digit from the sum of 1 . 2^1 + 2 . 2^2 + 3 . 2^3 + ... + 50 . 2^(50) ?


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Question Number 21738 by Joel577 last updated on 02/Oct/17

$What is the last digit from the sum of$ $1 . 2^{1} + 2 . 2^{2} + 3 . 2^{3} + . . . + 50 . 2^{50} ?$
	Answered by $@ty@m last updated on 02/Oct/17

	$S_{50} = 1 . 2^{1} + 2 . 2^{2} + 3 . 2^{3} + . . . + 50 . 2^{50}$ $2 . S_{50} = 1 . 2^{2} + 2 . 2^{3} + . . . . . . + 49 . 2^{50} + 50 . 2^{51}$ $______________________________$ $S u b t r a c t i n g$ $- S_{50} = 2^{1} + 2^{2} + 2^{3} + . . . . . . + 2^{50} - 50 . 2^{51}$ $- S_{50} = \frac{2 (2^{50} - 1)}{2 - 1} - 50 . 2^{51}$ $- S_{50} = 2^{51} - 2 - 50 . 2^{51}$ $S_{50} = 49 . 2^{51} + 2 - - - (1)$ $w e o b s e r v e t h a t$ $l a s t d i g i t i n 2^{1} = 2$ $l a s t d i g i t i n 2^{2} = 4$ $l a s t d i g i t i n 2^{3} = 8$ $l a s t d i g i t i n 2^{4} = 6$ $l a s t d i g i t i n 2^{5} = 2$ $.$ $.$ $. . . a n d s o o n .$ $\Rightarrow l a s t d i g i t i n 2^{4 n + 3} = 8$ $\Rightarrow l a s t d i g i t i n 2^{4 \times 12 + 3} = 8$ $\Rightarrow l a s t d i g i t i n 49 . 2^{51} + 2 = 4 A n s .$
	Commented by Joel577 last updated on 03/Oct/17

	$T h a n k s . B u t m y f r i e n d g o t 8$ $H e u s e d m o d t o f i n d t h e l a s t d i g i t$
	Commented by $@ty@m last updated on 03/Oct/17

	$I n t h a t c a s e e i t h e r h e o r I a m w r o n g .$ $T r y t o f i n d w h e r e i s t h e m i s t a k e .$
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