3 + 5 + 9 + 15 + 23 + …
Question:

3 + 5 + 9 + 15 + 23 + …

Solution:

Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum to $n$ terms of the given series.

Thus, we have:

$S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$    …(1)

Equation (1) can be rewritten as:

$S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n} \ldots(2)$

On subtracting (2) from (1), we get:

$S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$

$S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$

$0=3+\left[2+4+6+8+\ldots+\left(T_{n}-T_{n-1}\right)\right]-T_{n}$

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:

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

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

$\Rightarrow 3+n(n-1)=T_{n}$

Now,

$\because S_{n}=\sum_{k=1}^{n} T_{k}$

$\therefore S_{n}=\sum_{k=1}^{n}\{3+k(k-1)\}$

 

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 3-\sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{n(n+1)(2 n+1)}{6}+3 n-\frac{n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{n}{3}\left[\frac{(n+1)(2 n+1)}{2}+9-\frac{3}{2}(n+1)\right]$

$\Rightarrow S_{n}=\frac{n\left[n^{2}+8\right]}{3}$

 

 

 

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