Question

Find sum of n terms series :1×2×3+2×3×4+3×4×5+.....

Answer 1

The sum is:

S n

=

1.2

.3

+

2.3

.4

+

3.4

.5

+

...

+

n

( n + 1 )

( n + 2 )

=

n ∑ r = 1

r

( r + 1 )

( r + 2 )

=

n ∑ r = 1

r

( r 2 + 3 r + 2 )

= n ∑

r

=

1

(

r 3

+

3

r 2

+

2

r

)

=

n ∑ r = 1

r 3

+

3

n ∑ r = 1

r 2

+

2

n ∑ r = 1

r

We now use some standard formula for summation:

n ∑ r = 1

r

=

1 2

n

( n + 1 )

n ∑ r = 1

r 2

=

1 6

n

( n + 1 )

( 2 n + 1 )

n ∑ r = 1

r 3

=

1 4

n 2

( n + 1 ) 2

Which gives us:

S n

=

1 4

n 2

( n + 1 ) 2

+

3

⋅

1 6

n

( n + 1 )

( 2 n + 1 )

+

2

⋅

1 2

n

( n + 1 )

=

1 4

n 2

( n + 1 ) 2

+

1 2

n

( n + 1 )

( 2 n + 1 )

+

n

( n + 1 )

=

1 4

n

( n + 1 )

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

+

4

}

>

=

1 4

n

( n + 1 )

( n 2 + n + 4 n + 2 + 4 )

=

1 4

n

( n + 1 )

( n 2 + 5 n + 6 )

=

1 4

n

( n + 1 )

( n + 2 )

(