tìm số tự nhiên n sao cho 2 × $2^{2}$ + 3 × $2^{3}$ + 4 × $2^{4}$ +…….+n × $2^{n}$ = $2^{n+34}$

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1. Đáp án:

    

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2. Đáp án:

   2A = 2.$2^{3}$ + 3. $2^{4}$ + 4. $2^{5}$ +… + n. $2^{n+1}$   

   -A=A-2A =  2.$2^{2}$ + (3.$2^{3}$ – 2.$2^{3}$) + …. +( n-n+1) $2^{n}$  – n. $2^{n+1}$   

   = 2.$2^{2}$ + $2^{3}$ + $2^{4}$+ …. + $2^{n}$  – n. $2^{n+1}$   

   => A = -2.$2^{2}$ – ( $2^{2}$ +$2^{3}$ + $2^{4}$+ …. + $2^{n+1}$) + (n+1). $2^{n+1}$   

   B = $2^{2}$+$2^{3}$ + $2^{4}$ +  $2^{5}$ +… + $2^{n+1}$   

   2B = $2^{3}$ + $2^{4}$ +  $2^{5}$ +… + $2^{n+2}$ 

   B= 2B- B= $2^{n+2}$  – $2^{2}$

   => A = $2^{2}$  – $2^{n+2}$  -2.$2^{2}$ +(n+1). $2^{n+1}$

   =  (n+1). $2^{n+1}$ – $2^{n+2}$= (n+1-2). $2^{n+1}$= 2(n-1) $2^{n}$

   => 2(n-1) = $2^{34}$ => n = $2^{33}$ + 1

    

    

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