Question

find the first four terms of the sequences whose nth terms is a = (-1)(n+1)​

Answer 1

Answer:

tn = an = n3 - 2 (i) a1 = 13 – 2 = 1 – 2 – 1 a2 = 23 – 2 = 8 – 2 = 6 a3 = 33 – 2 = 27 – 2 = 25 a4 = 43 – 2 = 64 – 2 = 62 ∴ The first four terms are -1, 6, 25, 62, … (ii) an = (-1)n + 1 n(n + 1) a1 = (-1)1 + 1 (1) (1 +1) = (-1)2 (1) (2) = 2 a2 = (-1)2 + 1 (2) (2 + 1) = (-1)3 (2) (3) = -6 a3 = (-1)3 + 1 (3) (3 + 1) = (-1)4 (3) (4) = 12 a4 = (-1)4 + 1 (4) (4 + 1) = (-1)5 (4) (5) = -20 ∴ The first four terms are 2, -6, 12, -20,… (iii) an = 2n2 – 6 a1 = 2(1)2 – 6 = 2 – 6 = -4 a2 = 2(2)2 – 6 = 8 – 6 = 2 a3 = 2(3)2 – 6 = 18 – 6 = 12 a4 = 2(4)2 – 6 = 32 – 6 = 26 ∴ The first four terms are -4, 2, 12, 26Read more on Sarthaks.com - https://www.sarthaks.com/935293/find-the-first-four-terms-of-the-sequences-whose-nth-terms-are-given-by-i-an-n-3-2-ii-an-1-n-1-n-n-1