Question #217623

 For the given sequence 1/n, 1/2n, 1/4n,…  Find the 16th term. Find sum of first sixteen terms where n is your arid number i.e. 19-arid-234 take n=234


1
Expert's answer
2021-07-28T13:38:47-0400

Given sequence is

1n,12n,14n...\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}...

To find the 16th term and sum of first sixteen terms where n = 234.

We have the given sequence is

1n,12n,14n...\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}...

Now let us consider the terms of the sequence to be named as ak

Then

a1=1n,a2=12n,a3=14na_1= \frac{1}{n}, a_2 = \frac{1}{2n}, a_3=\frac{1}{4n}

Next

a2a1=12n1nanda3a2=14n12n=12\frac{a_2}{a_1}= \frac{ \frac{1}{2n} }{\frac{1}{n}} \\and \\\frac{a_3}{a_2}=\frac{\frac{1}{4n}}{\frac{1}{2n}}=\frac{1}{2} \\

hence the common ratio of the terms of the given sequence are equal so the given sequence is a geometric sequence whose first term is a1=1na_1 = \frac{1}{n} and common ratio r is r=12r=\frac{1}{2}

As the given sequence is a geometric sequence so its kth term is given by ak=a1rk1a_k=a_1r^{k-1} i.e ak=1n(12)k1a_k=\frac{1}{n}\Big(\frac{1}{2}\Big)^{k-1}

For the 16th term we consider k=16 in ak=1n(12)k1a_k=\frac{1}{n}\Big(\frac{1}{2}\Big)^{k-1} then we have

a16=1n(12)161a16=1n(12)15a16=132768na_{16}=\frac{1}{n}\Big(\frac{1}{2}\Big)^{16-1} \\ a_{16}=\frac{1}{n}\Big(\frac{1}{2}\Big)^{15} \\ a_{16}= \frac{1}{32768n}

Next the sum of first sixteen terms of the given sequence will be given by S16=a1(1r16)1rS_{16}=\frac{a_1(1-r^{16})}{1-r} as the formula for the sum of the first k terms with first term a1 and common ratio r<1 is given by Sk=a1(1rk)1rS_{k}=\frac{a_1(1-r^{k})}{1-r}

For a1=1na_1 = \frac{1}{n} and r=12r=\frac{1}{2} in S16=a1(1r16)1rS_{16}=\frac{a_1(1-r^{16})}{1-r} we get

S16=1n(1(12)16)112S16=1n(6553565536)12S16=6553532768nS_{16}=\frac{\frac{1}{n}(1-(\frac{1}{2})^{16})}{1-\frac{1}{2}} \\ S_{16}=\frac{\frac{1}{n}(\frac{65535}{65536})}{\frac{1}{2}} \\ S_{16}=\frac{65535}{32768n}

So the 16th term and sum of first sixteen terms of the sequence 1n,12n,14n\frac{1}{n}, \frac{1}{2n}, \frac{1}{4n}… are 132768n\frac{1}{32768n} and 6553532768n\frac{65535}{32768n} respectively.

As n = 234 so substituting n = 234 in the answers we get the 16th term to be 17667712\frac{1}{7667712} and sum of first sixteen terms to be 218452555904\frac{21845}{2555904}

Hence the required answers for n = 234 are 16th term is 17667712\frac{1}{7667712} and sum of first sixteen terms is 218452555904\frac{21845}{2555904}


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