Sequence there !

Algebra Level 3

What is the sum of first 25 terms of the sequence where n t h n^{th} term is defined by the formula

a n = n 2 − 2 n − 1 a_n = n^2 -2n -1

Details and assumptions

Sequence starts with a 1 a_1 , and it has a value a 1 = 1 − 2 − 1 = − 2 a_1 = 1-2-1=-2

a 2 = 2 2 − 2 × 2 − 1 = − 1 a_2 = 2^2-2\times 2 -1 = -1

a 3 = 3 2 − 3 × 2 − 1 = 2 a_3 = 3^2 - 3\times 2 -1 = 2

a 4 = 4 2 − 4 × 2 − 1 = 7 a_4 = 4^2-4\times 2 -1 =7 . . . .


The answer is 4850.

Aditya Raut by Aditya Raut , 23, India

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2 solutions

Aditya Raut
Jun 2, 2014

Clearly we can see that S n S_n (sum of n terms of the sequence) is

S n = ∑ k = 1 n k 2 − 2 ∑ k = 1 n k − n \displaystyle S_n = \sum_{k=1}^n k^2 -2\sum _{k=1}^{n} k - n

Thus S n = ( n ) ( n + 1 ) ( 2 n + 1 ) 6 − 2 n ( n + 1 ) 2 − n S_n= \dfrac{(n)(n+1)(2n+1)}{6} - 2 \dfrac{n(n+1)}{2} - n

Hence S 25 = 25 × 26 × 51 6 − 25 × 26 − 25 = 5525 − 650 − 25 = 4850 S_{25} = \dfrac{25\times 26\times 51}{6} -25\times 26 -25 =5525 - 650 -25 = \boxed{4850}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Same method

Bhargav Das - 7 years ago

class yfgubhjnk { public static void main(String args[]) { int s=0,i,n; for(i=1;i<=25;i++) { double j=(Math.pow(i,2)-2*i-1); int m=(int)j; s=s+m; } System.out.println("Sum= "+s); } }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Hey, what language programming is this? (I know only Python, and python can be written as if in latex, above is not in that form though)

Aditya Raut - 6 years, 3 months ago

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