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Question Number 214712 by efronzo1 last updated on 17/Dec/24
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Answered by mr W last updated on 17/Dec/24
r2−3r+2=0(r−1)(r−2)=0r=1,2⇒an=A×1n+B×2na0=A+B=2a1=A+2B=5⇒B=3,A=−1⇒an=3×2n−1✓
alternativeapproachan+2−an+1=2(an+1−an)saybn=an+1−anbn+1=2bn←G.P.⇒bn=2n×b0=2n(a1−a0)=3×2nan+1−an=3×2nan−an−1=3×2n−1......a1−a0=3×20an+1−a0=3(1+2+...+2n)=3×2n+1−12−1an+1=3×2n+1−12−1+2=3×2n+1−1⇒an=3×2n−1✓
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