Take it easy

Algebra Level 3

for any integer k>1000 A = ∑ n = 1 k ( − 1 ) n ( n + 1 ) 2 n + ∑ n = 1 k + 1 ( − 1 ) n ( n + 1 ) 2 n + ∑ n = 1 k + 2 ( − 1 ) n ( n + 1 ) 2 n + ∑ n = 1 k + 3 ( − 1 ) n ( n + 1 ) 2 n A= \large \displaystyle\sum_{n=1}^{k}(-1)^{\frac{n(n+1)}{2}}n + \large \displaystyle\sum_{n=1}^{k+1}(-1)^{\frac{n(n+1)}{2}}n + \large \displaystyle\sum_{n=1}^{k+2}(-1)^{\frac{n(n+1)}{2}}n + \large \displaystyle\sum_{n=1}^{k+3}(-1)^{\frac{n(n+1)}{2}}n if the sum of all possible values of A is m post your answer as 2m . note: don't combine the repeated values.


The answer is -8.

Hamza Omar by Hamza Omar , 23, Syria

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1 solution

Jafar Badour
Oct 29, 2015

if n ( m o d ( 4 ) ) = 0 o r ( n + 1 ) ( m o d ( 4 ) ) = 0 n(mod(4))=0 or (n+1)(mod(4))=0 then g ( n ) = n ( n + 1 ) 2 g(n)=\frac{n(n+1)}{2} is an even number this yields that H ( n ) = ( − 1 ) g ( n ) = 1 H(n)=(-1)^{g(n)} =1 else means H ( n ) = − 1 H(n) =-1 , its obvious that numbers are divisible by 4 take the sequence n=4,8,12,16,20...

so when n and n+1 are divisible by 4 that means the sequence is 4-1,4,8-1,8,12-1,12... meaning 3,4,7,8,11,12,...

when n takes the values above this means that g ( n ) g(n) is an even number which means H ( n ) = 1 H(n)=1

so when n takes 1 , 2, 3, 4, 5 ,6, 7, 8,9,10...

H(n) takes -1,-1,+1,+1,-1,-1,+1,+1,-1,-1...

I will simplify the equation

A = 4 ∑ i = 1 k + 3 f ( n + 1 ) + 2 f ( n + 2 ) + f ( n + 3 ) A=4\displaystyle \sum_{i=1}^k +3f(n+1)+2f(n+2)+f(n+3)

f ( n ) = ( − 1 ) n ( n + 1 ) 2 f(n)=(-1)^{\frac{n(n+1)}{2}}

∑ i = 1 k \displaystyle \sum_{i=1}^k while k>1000 then ∑ i = 1 k \displaystyle \sum_{i=1}^k is k ,0,-1,-k-1

if (k)(mod(4))=0 then ∑ i = 1 k = k \displaystyle \sum_{i=1}^k =k and

A = 4 k + 3 f ( n + 1 ) + 2 f ( n + 2 ) + f ( n + 3 ) A=4k+3f(n+1)+2f(n+2)+f(n+3)

A = 4 k − 3 ( k + 1 ) − 2 ( k + 2 ) + ( k + 3 ) = − 4 A=4k-3(k+1)-2(k+2)+(k+3)=-4

if (k+1)(mod(4))=0 then ∑ i = 1 k = 0 \displaystyle \sum_{i=1}^k =0 and

A = 0 + 3 ( k + 1 ) − 2 ( k + 2 ) − ( k + 3 ) = − 4 A=0+3(k+1)-2(k+2) -(k+3)=-4

if (k+2)(mod(4))=0 then ∑ i = 1 k = − k − 1 \displaystyle \sum_{i=1}^k =-k-1 and

A = − 4 − 4 k − 3 ( k + 1 ) + 2 ( k + 2 ) + ( k + 3 ) = 0 A=-4-4k-3(k+1)+2(k+2)+(k+3)=0

if (k)(mod(4))=1 then ∑ i = 1 k = − 1 \displaystyle \sum_{i=1}^k =-1 and

A = − 4 − 3 ( k + 1 ) + 2 ( k + 2 ) + ( k + 3 ) = 0 A=-4-3(k+1)+2(k+2)+(k+3)=0

the possible values are 0,-4 so m=-4 and 2m=-8!

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