Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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3 + 5 + 9 + 15 + 23 + ... - Mathematics

Short Note

3 + 5 + 9 + 15 + 23 + ...

 
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Solution

Let  \[T_n\]  be the nth term and \[S_n\]  be the sum to n terms of the given series. Thus, we have:  \[S_n = 3 + 5 + 9 + 15 + 23 + . . . + T_{n - 1} + T_n\]   ...(1) 

Equation (1) can be rewritten as: \[S_n = 3 + 5 + 9 + 15 + 23 + . . . + T_{n - 1} + T_n\]  ...(2) 

On subtracting (2) from (1), we get: 

\[S_n = 3 + 5 + 9 + 15 + 23 + . . . + T_{n - 1} + T_n \]
\[ S_n = 3 + 5 + 9 + 15 + 23 + . . . + T_{n - 1} + T_n \]
\[ 0 = 3 + \left[ 2 + 4 + 6 + 8 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n\]

The sequence of difference of successive terms is 2, 4, 6, 8,...
We observe that it is an AP with common difference 2 and first term 2.
Thus, we have:

\[3 + \left[ \frac{\left( n - 1 \right)}{2}\left\{ 4 + \left( n - 2 \right)2 \right\} \right] - T_n = 0\]
\[ \Rightarrow 3 + \left[ \frac{\left( n - 1 \right)}{2}\left( 2n \right) \right] = T_n \]
\[ \Rightarrow 3 + n\left( n - 1 \right) = T_n\]

Now,

\[\because S_n = \sum^n_{k = 1} T_k \]
\[ \therefore S_n = \sum^n_{k = 1} \left\{ 3 + k\left( k - 1 \right) \right\} \]
\[ \Rightarrow S_n = \sum^n_{k = 1} k^2 + \sum^n_{k = 1} 3 - \sum^n_{k = 1} k\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + 3n - \frac{n\left( n + 1 \right)}{2}\]
\[ \Rightarrow S_n = \frac{n}{3}\left[ \frac{\left( n + 1 \right)\left( 2n + 1 \right)}{2} + 9 - \frac{3}{2}\left( n + 1 \right) \right]\]
\[ \Rightarrow S_n = \frac{n\left[ n^2 + 8 \right]}{3}\]


 

  

 

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.2 | Q 1 | Page 18
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