Answer:
Explanation:
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 Sn .
∴Sn=1+3+(−5)+7+9+(−11)............................3n 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
∴Sn=(1+3+(−5)+7+9+(−11)...3n terms)+(5+11...n terms)−(5+11...n terms)
Now we will rearrange the equation according to our ease. On doing so, the equation becomes:
Sn=(1+3+5+7+9+11...3n terms)+((−5)+(−11)...n terms)−(5+11...n terms)
⇒Sn=(1+3+5+7+9+11...3n terms)−(5+11...n terms)−(5+11...n terms)
⇒Sn=(1+3+5+7+9+11...3n terms)−2(5+11...n terms)
So, we have divided Sn 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 k2(2a+(k−1)d) , where a is the first term of the arithmetic progression and d represents the common difference.
∴Sn=3n2(2×1+(3n−1)2)−2×n2(2×5+(n−1)6)
⇒Sn=3n×(1+3n−1)−×n(10+6n−6)
⇒Sn=3n×3n−n(4+6n)
⇒Sn=9n2−6n2−4n
∴Sn=3n2−4n
Therefore, the sum of the first 3n terms of the sequence mentioned in the question is equal to 3n2−4n
.