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Exercise 5.5 · Q12

Q.The nthn^{th} term of the sequence 1,2,4,7,11,⋯1,2,4,7,11,\cdots is

(1) n3+3n2+2nn^3+3n^2+2n
(2) n3−3n2+3nn^3-3n^2+3n
(3) n(n+1)(n+2)3\dfrac{n(n+1)(n+2)}{3}
(4) n2−n+22\dfrac{n^2-n+2}{2}
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Compute the first differences of the sequence — they form an AP — then sum them to get a closed formula for ana_n.

Step 1. Compute first differences. 2−1=1, 4−2=2, 7−4=3, 11−7=42-1=1,\ 4-2=2,\ 7-4=3,\ 11-7=4 — differences are 1,2,3,4,…1,2,3,4,\ldots, i.e. the difference before term nn is (n−1)(n-1).

Step 2. Sum the differences from the first term.

an=a1+∑k=1n−1k=1+(n−1)n2.a_n = a_1 + \sum_{k=1}^{n-1}k = 1+\frac{(n-1)n}2.

Step 3. Simplify to match the option form. …

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