a0=0,a1=212an+2=5an+1−3an2x2−5x+3=0⇒x=1,3/2∴an=Aln+B(23)n n=0n=10=A+B21=A+23 B]A=−1 B=1 ⇒an=−1+(23)nk=1∑100ak=k=1∑100(−1)+(23)k =−100+23−1(23)((23)100−1)=−100+3((23)100−1)=3⋅(a100)−100
Let ⟨an⟩ be a sequence such that a0=0,a1=21 and 2an+2=5an+1−3an,n=0,1,2,3,…. Then k=1∑100ak is equal to
Held on 28 Jan 2025 · Verified 6 Jul 2026.
3a99−100
3a100−100
3a99+100
3a100+100
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