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1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ... - Mathematics

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

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Solution

Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n = 1 + 2 + 3 + 4 + 5 + . . . + n = \frac{n\left( n + 1 \right)}{2} = \frac{n^2 + n}{2}\]

Now, let

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

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

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

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

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

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

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

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 4 \right)}{12}\]

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

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.1 | Q 5 | Page 10
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