Given sequence is
n1,2n1,4n1...
To find the 16th term and sum of first sixteen terms where n = 234.
We have the given sequence is
n1,2n1,4n1...
Now let us consider the terms of the sequence to be named as ak
Then
a1=n1,a2=2n1,a3=4n1
Next
a1a2=n12n1anda2a3=2n14n1=21
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=n1 and common ratio r is r=21
As the given sequence is a geometric sequence so its kth term is given by ak=a1rk−1 i.e ak=n1(21)k−1
For the 16th term we consider k=16 in ak=n1(21)k−1 then we have
a16=n1(21)16−1a16=n1(21)15a16=32768n1
Next the sum of first sixteen terms of the given sequence will be given by S16=1−ra1(1−r16) as the formula for the sum of the first k terms with first term a1 and common ratio r<1 is given by Sk=1−ra1(1−rk)
For a1=n1 and r=21 in S16=1−ra1(1−r16) we get
S16=1−21n1(1−(21)16)S16=21n1(6553665535)S16=32768n65535
So the 16th term and sum of first sixteen terms of the sequence n1,2n1,4n1… are 32768n1 and 32768n65535 respectively.
As n = 234 so substituting n = 234 in the answers we get the 16th term to be 76677121 and sum of first sixteen terms to be 255590421845
Hence the required answers for n = 234 are 16th term is 76677121 and sum of first sixteen terms is 255590421845
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