Question 6 - Finding sum from series - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Finding sum from series
Last updated at April 16, 2024 by Teachoo
Question 6, Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41… It is not an AP or a GP Let Sn = 5 + 11 + 19 + 29 + 41 ... + an–1 + an Sn = 0 + 5 + 11 + 19 + 41 ... + an–2 + an–1 + an Subtracting (2) from (1) Sn – Sn = 5 – 0 + [(11 – 5) + (19 – 11) + (29 – 19) + ...(an–1 – an–2 ) + (an – an – 1)] – an 0 = 5 + [6 + 8 + 10 + 12 + ...an–1 ] – an an = 5 + [6 + 8 + 10 + 12 + ... + (n – 1) terms] 6 + 8 + 10 + 12 + ... + (n – 1) term is an AP With first term a = 6 & common difference = d = 8 – 6 = 2 Sum of n terms of an AP = 𝑛/2 (2a + (n – 1)d) Putting n = n – 1 & a = 6 & d = 2 [6 + 8 + 10 + 12 + ... + (n – 1) terms] = (n−1)/2 [ 2(6) +((n – 1) – 1)2 ] = (n−1)/2 [ 12+(n – 1 – 1)2 ] = (n−1)/2 [ 12+(n – 2)2 ] = (n−1)/2 [12 + 2n – 4] = (n−1)/2 [8 + 2n] = (n−1)/2 × 2[4 + n] = (n – 1) (n + 4) Thus, [6 + 8 + 10+ … upto (n –1) terms] = (n – 1) (n + 4) Now, an = 5 + [6 + 8 + 10 + 12 + ... + (n – 1) terms] Putting values an = 5 + (n – 1)(n – 4) an = 5 + n(n + 4) – 1(n + 4) an = 5 + (n2 + 4n) – n – 4 an = 5 + n2 + 4n – n – 4 an = n2 + 3n + 1 Now = (n(n + 1)(2n + 1))/6 + 3((n(n + 1))/2) + n = (𝑛(n + 1)(2n + 1))/6 + 3/2 n(n + 1) + 𝑛/1 = (𝑛(n + 1)(2n + 1) + 9(n + 1)) + 6n)/6 = n(((n + 1)(2n + 1) + 9(n + 1) + 6)/6) = n((𝑛(2n + 1) + 1(2n + 1) + 9n + 9 + 6)/6) = n ((2n2 + 2n + n + 1 + 9n + 9 + 6)/6) = n((2n2 + 12n + 16)/6) = n((2(n2 +6n +8))/6) = n/3 (n2 + 6n +8) = n/3 [n(n + 4) + 2(n + 4)] = n/3 [(n + 2)(n + 4)] = (n(n + 2)(n + 4) )/3 Thus, the required sum is (n(n + 2)(n + 4) )/3