find the nth term of the following series
1,3,4,9,10,12,13......................
The above series is actually the increasing sequence of sum of numbers that are generated by sum of unique powers of 3
for example
4 = 1+3
10 = 1+9
12 = 3+9
13 = 1+3+9
find the nth term of the series
me and patro are not allowed to answer first
i want the explanation for this as this can be or may be done by more than one method
think about it its an easy question and also worth thinking for cat aspirants
ReplyDeleteans can be13,15,_,18,_ i.e alternate no.s are multiples of 3 and other alternates are 1,4,10,13....-->4->1+3
ReplyDelete10->4+6
13->10+3
next wud be 13+6 i.e 19
so series will go like 1,3,4,9,10,12,13,15,19...
sorry but 15 is not the sum of unique power of 3
ReplyDelete15 = 3+9+3
where 3 is repeating and thus cannot be expressed as unique power of 3
and the series after 13 will have 27
but a good way to explore series
nice one
but the question is not to explore series but to understand the given series and find the term
think a bit on that side and you will get
terms will be like
ReplyDelete27
1+27
3+27
1+3+27
9+27
1+9+27
1+3+9+27
if n=2^i then nth term will be 3^i
n+1 3^i+1
n+2 3^i+3
n+3 3^i+1+3
n+4 3^i+9
and like tat...
exact general term for nth term i dont know...
you are very to the solution just visualize what is the process and why you derived the above formula
ReplyDeleteyou will get your answer
f(n)=f(2^i)+f(n-2^i), where 2^i < n < 2^(i+1)
ReplyDeleteif n=2^i, then f(n)=3^i.
it iz like
ReplyDelete3^0=1
3^1=3
3^0+3^1=4
3^2=9
3^0+3^2=10
3^1+3^2=12
3^0+3^1+3^2=13
3^3=27
3^0+3^3=28
3^1+3^3=30
3^2+3^3=36
3^0+3^1+3^2+3^3=40
n similarly next wud be 3^4
n den adding dem wid powers of 3 individually uptill 3 n den all of dem uptill 4,,,,
so here pure power ie like only 3^0 or 1 or 2 wud come after n+1 terms between pure where n iz pure power...ie terms bw 3^3 and 3^4 wud be 4+1=5,n dere iz an exception uptill 3^2,,,ie dis rule starts from 3^2 wid n=2....from here if we hav 3^n (n+1)th term wud be 3^n+3^0,,
n+2th wud be wid 3^1...and n+nth wud be 3^0 +3^1+.....+3^n and then 3^n+1....
to Pankaj
ReplyDeletebut when does this nth term will be = 3^n
actually you need a relation on between the position of the number in the series and the value of the number
nth term will not be equal to 3^n
and actually you missed one of the terms in your series
3^3+3^1+3^0 = 31
To Anshul
Your answer is right but i need the way you reached the answer
value of nth term depends on value of 'n'
ReplyDeleteif n = 2^a + 2^b + 2^c +....
where 2^a <= n < 2^(a+1)
2^b < n - 2^a < 2^(b+1)
2^c < n -(2^a + 2^b)< 2^(c+1)
and so on
then 'nth term'= 3^a+3^b+3^c+...
this result was found by observing the series, as for every n=2^a, nth term=3^a
for every n=2^a+1=2^a+2^0,nth term=3^a+3^0
for every n=2^a+2=2^a+2^1,nth term=3^a+3^1
and so on
awesome reply by madhuri
ReplyDeleteNow its my turn to reply my interpretation
My interpretation or the final answer as same as madhuri's but i started the other way out
as we know that the series is a sum of unique powers of three
we can say that for creating a term we either select a particular power or not select it
a type of np complete or knapsack 01 problem
so what we get is the series is atually consisting of
a03^0+a13^1+a23^2---------------------;
where a0 and a1 and a2 are either 0 or 1
now the above expression is similar to the conversion of binary to decimal no
only 2 is replaced by 3
so my approach was if we want to get the value of n with unique powers of 2 we would convert it into its binary no and we could know that which no are selected or rejected
similarly here we need the unique powers of 3
so convert the nth term in its binary equivalent
now the binary string is multiplied by powers of 3 instead of 2 and we will be with our answer
now how the answer is similar to madhuris final answer
conversion of n to binary is equivalent to write a n in terms of
an2^n+an-12^n-1+an-22^n-2---------------a02^0
where ax is either 0 or 1
so the above expression can be written as
2^a+2^b+2^c------------
where a,b,c are not equal
no multiplying the coefficients by correspondin power of 3
so the corresponding nth term will be
an3^n+an-13^n-1+an-23^n-2---------------a03^0
or
3^a+3^b+3^c-------and so on
similarly the anshuls answer can be traced of
by recursive subtraction
if n is considered to be sum unique powers of 2
(binary equivalent)
then recursvely or iteratively if we start with i = 0;
so if 2^0 exists in the term it is substituted with 3^0 and subtracted from initial no this process is repeated again and again
and we get the anshuls answer