Question 11 - 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
Misc 23 Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + … It is not an AP or a GP Let Sn = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an Sn = 0 + 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an Subtracting (2) from (1) Sn – Sn = 3 – 0 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an-1 – an – 2) + (an – an – 1)] – an 0 = 3 + [4 + 6 + 8 + … an –1 ] – an an = 3 + [4 + 6 + 8 + … + an –1 ] Now 4 + 6 + 8 + … + an – 1 is an AP Whose first term a = 4 common difference d = 6 – 4 = 2 We know that, Sum of n terms of AP = 𝑛/2 (2a + (n – 1)d) Putting n = n – 1 , a = 4 , d = 2 [4 + 6 + 8 + … (n –1) terms] = ((n − 1)/2) [ 2a +(n – 1 – 1)d] = ((n − 1)/2) [ 2(4) +(n – 2)2 ] = ((n − 1)/2) [ 8 + 2n – 4 ] = ((n − 1)/2) [ 2n + 4 ] = (n−1)/2 × 2(n+2) = (n – 1) (n + 2) Thus, [4 + 6 + 8 + … upto (n –1) terms] = (n – 1) (n + 2) From (3) an = 3 + [4 + 6 + 8 + … + an –1 ] Putting values = 3 + (n – 1) (n + 2) = 3 + n(n + 2) – 1(n + 2) = 3 + n2 + 2n – n – 2 = n2 + n + 3 – 2 = n2 + n + 1 Now = (n(n + 1)(2n + 1))/6 + (n(n + 1))/2 + n = (n(n + 1)(2n + 1) + 3n(n + 1) + 6n)/6 = n(((n + 1)(2n + 1) + 3(n + 1) + 6)/6) = n ((2n2 + n + 2n + 1 + 3n + 3 + 6)/6) = n ((2n2+6n+10)/6) = n/6 × 2 (n2 + 3n +5) = n/3 (n2 + 3n +5) Thus, the required sum is n/3 (n2 + 3n +5)