Working from the end

Algebra Level 3

( 1 2 − t 1 ) + ( 2 2 − t 2 ) + . . . + ( n 2 − t n ) = 1 3 n ( n 2 − 1 ) (1^2-t_1)+(2^2-t_2)+...+(n^2-t_n)=\dfrac{1}{3} n(n^2-1)

If { t n } \{t_n\} be a sequence of the terms satisfying the above equation, then find the 172 9 t h 1729^{th} term of the sequence.

You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.


The answer is 1729.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

It is given that:

∑ k = 1 n ( k 2 − t k ) = 1 3 n ( n 2 − 1 ) = 1 3 n ( n − 1 ) ( n + 1 ) \begin{aligned} \sum_{k=1}^n {(k^2-t_k)} & = \frac{1}{3}n(n^2-1) = \frac{1}{3}n(n-1)(n+1) \end{aligned} .

Therefore, the n t h n^{th} term is given by:

n 2 − t n = 1 3 n ( n − 1 ) ( n + 1 ) − 1 3 ( n − 1 ) ( n − 2 ) ( n ) = 1 3 n ( n − 1 ) ( n + 1 − n + 2 ) = n ( n − 1 ) ⇒ n 2 − t n = n 2 − n ⇒ t n = n ⇒ t 1729 = 1729 \begin{aligned} n^2-t_n & = \frac{1}{3}n(n-1)(n+1) - \frac{1}{3}(n-1)(n-2)(n) \\ & = \frac{1}{3}n(n-1)(n+1-n+2) \\ & = n(n-1) \\ \Rightarrow n^2-t_n & = n^2 - n \\ \Rightarrow t_n & = n \\ \Rightarrow t_{1729} & = \boxed{1729} \end{aligned}

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Moderator note:

Simple standard approach of determining each t n t_n by induction.

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