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Help with this question asap!!

Help with this question asap!!-example-1
User MacNimble
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1 Answer

3 votes

Answer:

1st option.


\sum_(k=1)^(6) (1)/(2k-1)

Explanation:

The given sequence is as follows:


1 + (1)/(3)+ (1)/(5)+ (1)/(7)+ (1)/(9)+ (1)/(11)

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:


a_n = a+(n-1)d

where a is the first term and

d is the common difference.

So, nth term for above AP:


a_n = 1+(n-1)* 2\\\Rightarrow 2n-1

Here, we have 'k' in place of 'n':


a_k = 2k-1

And as per the given sequence in the question, the kth term is simply the reciprocal:

i.e.
(1)/(2k-1)

and sum is to be done up to k = 6 from k = 1.

Hence, the summation notation for the sequence


1 + (1)/(3)+ (1)/(5)+ (1)/(7)+ (1)/(9)+ (1)/(11)

is:


\sum_(k=1)^(6) (1)/(2k-1)

User Jernej Novak
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