The sequence of real numbers b1, b2, b3, ..., is such that b1 = 4 and bn = bn-1 + 2n for all n ∈ Z+, n≥2. Using the method of difference, show that bn = n2 + n + 2 for all n ∈ Z+.
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Answer:
Using Method of difference,
bn - b1
= bn - bn-1
+ bn-1 - bn-2
+ bn-2 - bn-3
+ bn-3 - bn-4
+ ..... - .....
+ b3 - b2
+ b2 - b1
= 2(2) + 2(3) + ... + 2(n-2) + 2(n-1) + 2(n)
= an AP with first term 4, last term 2n, and (n-1) terms
= (n-1)/2 * (4 + 2n)
= (n-1)(n+2)
= n2 + n - 2
So,
bn - 4 = n2 + n - 2
bn = n2 + n + 2 (shown)