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Consider the sequence of sums 13,13+19,13+19+127,13+19+127+181,

(a) What happens to the terms of this sequence of sums as k gets larger and larger?

(b) Find a sufficiently large value of k which will guarantee that every term past the kth term of this sequence of sums is in the interval (0.49999, 0.5).

Short Answer

Expert verified

(a)limk13k+13k1+13k2++13=

(b)0.4974

Step by step solution

01

Part (a) Step 1. Given information

Given is the sequence 13,13+19,13+19+127,13+19+127+181,

We have to explain what happens to the terms of this sequence as k gets larger and larger and find a sufficiently large value of k which will guarantee that every term past the kth term of this sequence of sums is in the interval (0.49999, 0.5).

02

Part (a) Step 2. Terms of the sequence

From the sequence, we see that, as k gets larger and large, the terms get larger and larger.

Therefore, in limit expression it can be written as below:

limk13k+13k1+13k2++13=

03

Part (b) Step 1. Value of k

For every k>5, the terms will be :

135+1351+1352+1353+13=135+134+133+132+130.00412+0.0123+0.037+0.111+0.3330.4974

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