∴3*2+3d=12, 답은 d=2입니다.
∴a[n]=2+2(n-1)=2n
(2)∵b[n]=a[n]2^a[n)
∴b[n]? =2n2^(2n)=2n4^n
∴t[n]/2=1*4^1+2*4^2+3*4^3+... +(n-1)4^(n-1)+n4^n
∵4t[n]/2=1*4^2+2*4^3+3*4^4... +(n-1)4^n+n4^(n+1)
∴3T[n]/2
= 4T n/2-T n/2
=n4^(n+1)-(4^1+4^2+4^3+...+...
=n4^(n+1)-(4^1+4^2+...+4^3...
. +4^n)
=n4^(n+1)-4(4^n-1)/(4-1)
=n4^(n+1)-4(4^n-1)/3
∴t[n]=2n4^(n+1)/3-8(4^n-1)/9
=6n4^(n+1)/9-[2*4^(n+1)-8 ]/9
=[(6n-2)4^(n+1)+8]/9
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작은 루핑 프로그램을 컴파일하고 다음을 확인합니다:
a[1]=8, a[2]=72, a[3]=456, a[4]=2504, a[5]=12744
a[6]=61896. a[7]=291272, a[8]=1339848
a[9]=6058440, a[10]=27029960, ......
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