Question

The sum of the series 1.2.3+2.3.4+3.4.5+.......To n terms is

Answer 1

Answer:

Step-by-step explanation:

tn=n(n+1)(n+2)

tn can be written as

1/4(n(n+1)(n+2)(n+3)-(n-1)(n)(n+1)(n+2))

it is n the form

vn-vn-1

substitute 1 in vn-1 and n in vn

n(n+1)(n+2)(n+3)-0

n(n+1)(n+2)(n+3)=sn

sum of terms =1/4(n(n+1)(n+2)(n+3))

Answer 2

$\huge{\orange{\boxed{\boxed{\boxed{\purple{\underline{\underline{\red{\mathfrak{Hey, mate...!!!}}}}}}}}}}$$\huge{\underline{\green{\mathbf{Answer:-}}}}$

✧══════•❁❀❁•══════✧

✧══════•❁❀❁•══════✧