Question Number 102539 by Dwaipayan Shikari last updated on 09/Jul/20

2+3.3+4.3^2 +5.3^2 +.....up to n terms

Commented byDwaipayan Shikari last updated on 09/Jul/20

Thanking both of you

Commented byRasheed.Sindhi last updated on 10/Jul/20

2+3.3+4.3^2 +5.3^3 +.....up to n  terms

Answered by PRITHWISH SEN 2 last updated on 09/Jul/20

t_n = (n+1)3^(n−1) = n3^(n−1) +3^(n−1)   t_1 = 1.3^0 + 3^0   t_2  = 2.3^1 + 3^1   t_3 = 3.3^2 +3^2   ..................  t_n = n.3^(n−1) +3^(n−1)   S_n = ((1−(1+n)3^n )/(1−3))+((3(1−3^n ))/((1−3)^2 )) + ((3^n −1)/2)    = 3^n (n/2)+(3^n /4)−(1/4)     please check.

Answered by mr W last updated on 09/Jul/20

S_n =Σ_(k=1) ^n (k+1)3^(k−1)   S_n =Σ_(k=1) ^n k3^(k−1) +Σ_(k=1) ^n 3^(k−1)   S_n =Σ_(k=0) ^(n−1) (k+1)3^k +Σ_(k=1) ^n 3^(k−1)   S_n =Σ_(k=1) ^n (k+1)3^k +1−(n+1)3^n +Σ_(k=1) ^n 3^(k−1)   S_n =3Σ_(k=1) ^n (k+1)3^(k−1) +1−(n+1)3^n +Σ_(k=1) ^n 3^(k−1)   S_n =3S_n +1−(n+1)3^n +((3^n −1)/(3−1))  ⇒S_n =(((2n+1)3^n −1)/4)

Commented byDwaipayan Shikari last updated on 09/Jul/20

Thanking both of you

Answered by Dwaipayan Shikari last updated on 09/Jul/20

S_n =2+3.3+4.3^2 +.......+(n+1)3^(n−1)   3S_n =    2.3+3.3^2 +...........+n3^(n−1) +3(n+1)3^(n−1)   ...............subtracting  −2S_n =2+3+3^2 +3^3 +.....3^(n−1) −3(n+1)3^(n−1)   S_n =(3/2)(n+1)3^(n−1) −(1/2)(2+3+3^2 +....n)  S_n =(1/2)(n+1)3^n −(1/2)(2+((3^n −1)/2))  S_n =(((2n+2)3^n −4−3^n +3)/4)=(((2n+1)3^n −1)/4)