Question

Question 8 Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Class X1 - Maths -Sequences and Series Page 196

Answer 1

n(n+1)(n+4)
n^2+n(n+4)
n^3+4n^2+n^2+4n
n^3+5n^2+4n

Answer 2

Tn = n(n + 1)( n + 4)
we know,
Sn = ∑Tn
=∑{n(n + 1)(n + 4)}
=∑{n³ + 5n² + 4n }
=∑n³ + 5∑n² + 4∑n

we know ,
∑n³ = [n(n + 1)/2]²
∑n² = n(n + 1)(2n + 1)/6
∑n = n( n + 1)/2 , use this here,

Sn = [n(n + 1)/2]² + 5n(n +1)(2n+1)/6 + 4n(n+1)/2
= n(n +1)/2[ n(n+1)/2 + 5(2n +1)/3 + 4]
= n(n + 1)/2[ {3n² + 3n + 20n + 10 + 24}/6]
= n(n+1)/12[3n² + 23n + 34]
= n(n+1)/12 [ 3n² + 6n + 17n + 34]
= n(n +1)/12 (3n + 17)(n+2)
= n(n+1)(n+2)(3n+17)/12