Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ... - Mathematics

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

Advertisement Remove all ads

Solution

Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n  = n\left( n + 1 \right) =  n^2  + n\]

Now, let \[S_n\] be the sum of n terms of the given series.

Thus, we have: \[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum^n_{k = 1} \left( k^2 + k \right)\]

\[ \Rightarrow S_n = \sum^n_{k = 1} k^2 + \sum^n_{k = 1} k\]

\[ \Rightarrow S_n = \left( \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{n\left( n + 1 \right)}{2} \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{2n + 1}{3} + 1 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{2n + 4}{3} \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 4 \right)}{6}\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( n + 2 \right)}{3}\]

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.1 | Q 6 | Page 10
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×