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If 1 + 1 + 2 2 + 1 + 2 + 3 3 + . . . . to N Terms is S, Then S is Equal to - Mathematics

MCQ

If \[1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .\] to n terms is S, then S is equal to

Options

  • \[\frac{n (n + 3)}{4}\]

  • \[\frac{n (n + 2)}{4}\]

  • \[\frac{n (n + 1) (n + 2)}{6}\]

  •  n2

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Solution

\[\frac{n (n + 3)}{4}\]

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = \frac{1 + 2 + 3 + 4 + 5 + . . . + n}{n} = \frac{n\left( n + 1 \right)}{2n} = \frac{n}{2} + \frac{1}{2}\]

Now, let

\[S_n\]  be the sum of n terms of the given series.
Thus, we have:

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

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

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

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

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

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

  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
Q 6 | Page 20
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