Question

1+3+6+10+............+n terms =1/6 n (n+1) (n+2)​

Answer 1

Answer:

ANSWER

Let S=1+3+6+10+..............+T

n

........(1)

Also S=1+3+6+10+..........+T

n−1

+T

n

........(2)

Now by (1)−(2)

0=1+2+3+.........+n−T

n

⇒T

n

=1+2+3......+n=

2

1

n(n+1)

Hence required summation is,

k=1

∑

n

T

k

=

2

1

(

k=1

∑

n

k

2

+

k=1

∑

n

k)

=

2

1

(

6

1

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

2

1

n(n+1))

=

6

1

n(n+1)(n+2)