How to do it?

We note that all the even terms form a series and all the odd terms form another series. Now the $1729^{th}$ term belongs to the series formed by the odd terms(why ???) .

Now we have to find the relation between the number of terms of the odd series and the original series, which is : Every $n^{th}$ odd term is the $(2n-1)^{th}$ term in the original series .

let $T_n =T_{1729}$ , therefore it is also the $T_{2n-1}$ of the odd series ,

$2n-1 =1729$ ; $n=865$

Now the odd series is actually an A.P with the first term = 1 and common difference = 5.

Therefore , $T_{865} =1 + 864 \times (5-1)$ ; $T_{865} =4321$ .

I noticed that every odd term was 3 more than the previous term and every even term was 2 more than the previous term. So I found that on average every term was 2.5 more than the previous term. Using this, I made the sequence ${2.5n-1.5}$ and substituted ${1729}$ as ${n}$ .

${2.5(1729)-1.5 = \boxed{4321}}$

difference changes frequently for odd and even terms by 2 and 3 respectively since 1729 is odd , take only odd terms 1,6,11,... (with difference 5) and the term gets reduced to (1729+1/2) i.e 865 then apply AP formula to find 865 th term 1+(865-1)*5 =4321

Use 1+(864×2)+(864×3)=4321

Why is it tagged under calculus

Note that the general expression for the series may been expressed as:-

$a_{n}=2n+\lfloor{\frac{n-3}{2}}\rfloor$

Thus: $a_{1729}=2(1729)+\lfloor{\frac{1729-3}{2}}\rfloor =\fbox{4321}$

Looking at the sequence above we can deduce that the sequence goes as follows:

............, +2, +3, +2, +3, ..............

Now, since the difference between any given two terms is not uniform, it would imply that the formula for obtaining the number at an odd position would be different from the formula for obtaining the number at an even position. Since the question asks for 1729th term , we can focus on the numbers at odd positions alone .

Therefore, with regards to the odd positions, notice the following, where f(n) is f(number at position) :

f(3) = 6 = 2*3

f(5) = 11 = 2*5 + 1

f(7) = 16 = 2*7 + 2

f(9) = 21 = 2*9 + 3

f(11) = 26 = 2*11 + 4

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This can be generalised as:

f(n) = 2n + (n-3)/2

Leading us to the solution for f(1729) : = 2*1729 + (1726/2)

= 4321

I'm not sure the formula used to solve this problem. Just thought of some relationship between these numbers. Writing it below.

If you note the sequence, each cycle will have 4 values (i.e) the first cycle has 1,3,6,8 and the second cycle has 11,13,16,18 and the third has 21,23,26,28 and so on...

So, If we divide the nth position by 4, we will get the number of cycle. Multiply the number of completed cycle (i.e) quotient and add the value in the remainder's position will give the value in the nth position.

The equation for this would be, Quotient*10+value in the remainder position from the cycle.

For example,

If we want to know the value in the 5th position, divide 5 by 4, you will get 1 quotient(1st cycle) and remainder 1. So, the value will be at first position in the second cycle. (i.e) 1*10+1=11

Likewise, to know 15th position, which will be on 4th cycle (15 divided by 4,give quotient as 3 and the remainder as 3, which surely lies on the 4th cycle). So the value can be calculated as 3*10+6=36.

In the same way, for the value in 1729th position,

divide 1729 by 4, which gives 432 and remainder as 1. So now, multiply 10 with the completed circle number (i.e) 432 and add the values in the remainder's position (i.e) 1.

432*10+1=4320+1=4321 will be at 1729th position.

\left( \frac { n-1 }{ 2 } \right) \quad \times \quad 5\quad +\quad 1\quad