Toán Lớp 7: Tìm stn n thỏa mãn điều kiện : 2.2² + 3.2 ³ + 4. $2^{4}$ + … + n. $2^{n}$ = $2^{n + 11}$
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Toán Lớp 7: Tìm stn n thỏa mãn điều kiện : 2.2² + 3.2 ³ + 4. $2^{4}$ + … + n. $2^{n}$ = $2^{n + 11}$
Comments ( 2 )
Ta có:
A=2.22+3.23+4.24+…+n.2nA=2.22+3.23+4.24+…+n.2n
⇒2A=2(2.22+3.23+4.24+…+n.2n)⇒2A=2(2.22+3.23+4.24+…+n.2n)
⇒2A=2.23+3.24+4.25+…+n.2n+1⇒2A=2.23+3.24+4.25+…+n.2n+1
⇒2A−A=2.22+(3.23−2.23)+…+(n−n+1).2n−n.2n+1⇒2A−A=2.22+(3.23−2.23)+…+(n−n+1).2n−n.2n+1
⇒A=2.22+23+24+…+2n−n.2n+1⇒A=2.22+23+24+…+2n−n.2n+1
⇒A=22+(22+23+…+2n+1)−(n+1).2n+1⇒A=22+(22+23+…+2n+1)−(n+1).2n+1
⇒A=−22−(22+23+…+2n+1)+(n+1).2n+1⇒A=−22−(22+23+…+2n+1)+(n+1).2n+1
Đặt B=22+23+…+2n+1B=22+23+…+2n+1
⇒2B=23+24+…+2n+2⇒2B=23+24+…+2n+2
⇒2B−B=2n+2−22⇒B=2n+2−22⇒2B−B=2n+2−22⇒B=2n+2−22
⇒A=22−2n+2+22+(n+1).2n+1⇒A=22−2n+2+22+(n+1).2n+1
⇒A=(n+1).2n+1−2n+2⇒A=(n+1).2n+1−2n+2
⇒A=2n+1(n+1−2)⇒A=2n+1(n+1−2)
⇒A=(n−1).2n+1=2(n−1).2n⇒A=(n−1).2n+1=2(n−1).2n
Mà A=2(n−1).2n=2n+10A=2(n−1).2n=2n+10
⇒2(n+1)=210⇒n−1=29⇒2(n+1)=210⇒n−1=29
⇒n−1=512⇒n=513⇒n−1=512⇒n=513
Vậy n=513