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In an arithmetic progression whose nth term is a_n, a_1 = 6, a_2 = 8, and a_3 = 10.  A second sequence, whose nth term is b_n, has b_1 = 3, and for n > 1, b_n = b_{n - 1} + a_{n - 1}.  Write an equation expressing b_n explicitly in terms of n.

 Dec 8, 2019
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Observing the recurrence relation, \(\{b_n\} \) is a quadratic sequence.

 

Let \(b_n = pn^2 + qn + r\qquad p,q,r\in \mathbb R\)

 

Obviously, \(a_n = 4 + 2n\).

 

\(b_1 = 3\\ p(1)^2 + q(1) + r = 3\\ p + q +r = 3 \phantom{-----}(1)\)

 

\(b_n = b_{n - 1} + a_{n - 1}\\ pn^2 + qn + r = p(n - 1)^2 + q(n - 1) + r +(4+2(n-1))\\ pn^2 +qn + r = pn^2 - 2pn + p + qn - q + r + 4 + 2n - 2\\ 0=-2pn+p-q+2+2n\\ (2 - 2p)n+(p-q+2)=0n+0\\ \text{Comparing both sides,}\\ 2-2p = 0\\ 2p = 2\\ \boxed{p = 1}\\ p -q+2=0\\ 1-q+2=0\\ \boxed{q=3}\\ \text{Substitute }p=1\text{ and }q = 3\text{ into equation }(1),\\ 1+3+r=3\\ \boxed{r = -1}\\ \therefore b_n = n^2 + 3n - 1\)

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 Dec 8, 2019
edited by MaxWong  Dec 8, 2019

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