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

n (n + 1) (n + 2)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(n + 1) (n + 2) (n + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4} n (n + 1) (n + 2) (n + 3)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{4} (n + 1) (n + 2) (n + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $n (n + 1) (n + 2)$
C $\frac{1}{4} n (n + 1) (n + 2) (n + 3)$

Suggest Corrections

Similar questions

1.2.3 + 2.3.4 + 3.4.5 + . . . n (n + 1) (n + 2)

\frac{n (n + 1) (n + 2) (n + 3)}{4}

Using the principle of mathematical induction, prove that

$1.2.3 + 2.3.4 + 3.4.5 + . . . + n (n + 1) (n + 2) = \frac{n (n + 1) (n + 2) (n + 3)}{4},$ for all $n \in N .$

1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \dots + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)} .$

\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + . . . . . + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)}

Join BYJU'S Learning Program