Question

2.3.1+3.4.4+4.5.7+....n terms=

Answer 1

Answer:

General term : (1+k)(2+k)(1+3k) where k=1 to n

∑

k=1

n

(1+k)(2+k)(1+3k)=∑

k=1

n

3k

3

+10k

2

+9k+2

=3

4

n

2

(n+1)

2

+10

6

(n)(n+1)(2n+1)

+9

2

(n)(n+1)

+2n

=

12

n(9n

3

+64n

2

+15n−34)