 QUESTION

# Sum up to 3n terms of the sequence 1, 3, -5, 7, 9, -11, 13, 15, -17…………. is.

Hint: Notice the pattern and try to separate the sum of the given sequence in terms of the sum of two different arithmetic progressions.

The given sequence is 1, 3, -5, 7, 9, -11, 13, 15, -17…………. . In the given sequence, the absolute values of the terms are in arithmetic progression with a negative number appearing after every two positive terms of the sequence.
Now moving to find the sum of the arithmetic progression. The sum of a sequence is generally denoted by ${{S}_{n}}$ .
$\therefore {{S}_{n}}=1+3+\left( -5 \right)+7+9+\left( -11 \right)............................3n\text{ terms}$
Out of the 3n terms, n terms are negative, while 2n terms are positive in the above sequence.
Now adding and subtracting the negative terms in our equation, we get
$\therefore {{S}_{n}}=\left( 1+3+\left( -5 \right)+7+9+\left( -11 \right)...3n\text{ terms} \right)\text{+}\left( \text{5+11}...\text{n terms} \right)-\left( \text{5+11}...\text{n terms} \right)$
Now we will rearrange the equation according to our ease. On doing so, the equation becomes:
${{S}_{n}}=\left( 1+3+5+7+9+11...3n\text{ terms} \right)\text{+}\left( \left( -5 \right)\text{+}\left( -11 \right)...\text{n terms} \right)-\left( \text{5+11}...\text{n terms} \right)$
$\Rightarrow {{S}_{n}}=\left( 1+3+5+7+9+11...3n\text{ terms} \right)-\left( \text{5+11}...\text{n terms} \right)-\left( \text{5+11}...\text{n terms} \right)$
$\Rightarrow {{S}_{n}}=\left( 1+3+5+7+9+11...3n\text{ terms} \right)-2\left( \text{5+11}...\text{n terms} \right)$
So, we have divided ${{S}_{n}}$ in two different series, and each series is an arithmetic progression. Now we know that sum of k terms of an arithmetic progression is equal to $\dfrac{k}{2}\left( 2a+(k-1)d \right)$ , where a is the first term of the arithmetic progression and d represents the common difference.
$\therefore {{S}_{n}}=\dfrac{3n}{2}\left( 2\times 1+\left( 3n-1 \right)2 \right)-2\times \dfrac{n}{2}\left( 2\times 5+\left( n-1 \right)6 \right)$
$\Rightarrow {{S}_{n}}=\dfrac{3n}{{}}\times \left( 1+3n-1 \right)-\times \dfrac{n}{{}}\left( 10+6n-6 \right)$
$\Rightarrow {{S}_{n}}=3n\times 3n-n\left( 4+6n \right)$
$\Rightarrow {{S}_{n}}=9{{n}^{2}}-6{{n}^{2}}-4n$
$\therefore {{S}_{n}}=3{{n}^{2}}-4n$
Therefore, the sum of the first 3n terms of the sequence mentioned in the question is equal to $3{{n}^{2}}-4n$ .

Note: Be careful about the signs and try to keep the equations as neat as possible by removing the removable terms. Moreover, make sure that you learn the formulas related to arithmetic progressions and geometric progressions as they are used quite often.