Answer:
1st option.
[tex]\sum_{k=1}^{6} \frac{1}{2k-1}[/tex]
Step-by-step explanation:
The given sequence is as follows:
[tex]1 + \dfrac{1}{3}+ \dfrac{1}{5}+ \dfrac{1}{7}+ \dfrac{1}{9}+ \dfrac{1}{11}[/tex]
Here, we can see that the denominator is in AP and the sequence is:
1, 3, 7, 9, 11
first term, a = 1
Common difference, d = 2
We know that nth term for an AP:
[tex]a_n = a+(n-1)d[/tex]
where a is the first term and
d is the common difference.
So, nth term for above AP:
[tex]a_n = 1+(n-1)\times 2\\\Rightarrow 2n-1[/tex]
Here, we have 'k' in place of 'n':
[tex]a_k = 2k-1[/tex]
And as per the given sequence in the question, the kth term is simply the reciprocal:
i.e. [tex]\dfrac{1}{2k-1}[/tex]
and sum is to be done up to k = 6 from k = 1.
Hence, the summation notation for the sequence
[tex]1 + \dfrac{1}{3}+ \dfrac{1}{5}+ \dfrac{1}{7}+ \dfrac{1}{9}+ \dfrac{1}{11}[/tex]
is:
[tex]\sum_{k=1}^{6} \frac{1}{2k-1}[/tex]
