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Question Number 49036 by Tinkutara last updated on 01/Dec/18

Commented by prakash jain last updated on 02/Dec/18

$2^{n - 1} g a v e a t l e a s t 1 w r o n g a n s w e r$ $(i n c l u d e s t u d e n t w i t h 2 w r o n g a n s w e r s)$ $2^{n - 2} g a v e a t l e a s t 2 w r o n g a m s w e r s$ $(o n l y o n e a d d i t i o n a l w r o n g a n s w e r$ $s h o u l d b e c o u n t e d)$ $. . .$ $2^{n - k} g a v e k w r o n g a n s w e r$ $S = 2^{n - 1} + 2^{n - 2} + . . . + 2^{n - k}$ $2^{n - 1} (\frac{1 - {(\frac{1}{2})}^{k}}{1 - \frac{1}{2}}) = 2^{n} \times \frac{2^{k} - 1}{2^{k}}$ $e x a m p l e$ $n = 6, k = 3$ $32 - 1 o r m o r e w r o n g$ $16 - 2 o f m o r e w r o n g$ $8 - 3 o r m o r e w r o n g$ $t o t a l w r o n g a n s w e r 32 + 16 + 8 = 56$
Commented by prakash jain last updated on 02/Dec/18

$Book answer will be correct if n = k .$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18

$t h a n k y o u s i r . . .$
Commented by Tinkutara last updated on 03/Dec/18
Thank you very much Sir! I got the answer. ��
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

	$p u t i = 1$ $2^{n - 1} s t u d e n t s g i v e 1 w r o n g a n s w e r$ $2^{n - 2} s t u d e n t s g i v e 2 w r o n g a n s r w e r$ $2^{n - 3} s t u d e n t s g i v e 3 w r o n g a n s w e r$ $. . .$ $. . .$ $2^{n - k} s t u d e n t s g i v e k w r o n g a n s w e r$ $t o t a l w r o n g a n s w e r$ $S = 2^{n - 1} \times 1 + 2^{n - 2} \times 2 + 2^{n - 3} \times 3 . . . + 2^{n - k} \times k$ $S \times 2^{- 1} = 2^{n - 2} \times 1 + 2^{n - 3} \times 2 . . + 2^{n - k} \times (k - 1) + 2^{n - k - 1} \times k$ $s u b s t r u c t$ $(S - \frac{S}{2}) = (2^{n - 1} + 2^{n - 1} + . . . + 2^{n - 1} n t i m e s) - 2^{n - k - 1} \times k$ $\frac{S}{2} = n 2^{n - 1} - 2^{n - k - 1} \times k$ $S = n 2^{n} - 2^{n - k} \times k$
	Commented by Tinkutara last updated on 02/Dec/18
	But answer is independent of k.
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

	$0 k l e t o t h e r s t r y o r y o u p o s t a n s w e r i n d e t a i l s$
	Commented by Tinkutara last updated on 02/Dec/18

	$A n s i s 2^{n} - 1$ $I {d o n}^{'} t u n d e r s t a n d t h e s o l u t i o n g i v e n$
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