mth root of unity?

Geometry Level 5

Given that A k = k ( k − 1 ) 2 cos ⁡ k ( k − 1 ) π 2 A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2

find ∣ A 19 + A 20 + ⋯ + A 98 ∣ . |A_{19} + A_{20} + \cdots + A_{98}|.

Try my set


The answer is 40.

Parth Lohomi by Parth Lohomi , 20, India

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2 solutions

Chew-Seong Cheong
Apr 15, 2015

A k = k ( k − 1 ) 2 cos ⁡ k ( k − 1 ) π 2 = { + k ( k − 1 ) 2 if k ( k − 1 ) 2 is even − k ( k − 1 ) 2 if k ( k − 1 ) 2 is odd A_k = \dfrac {k(k-1)}{2} \cos{\dfrac{k(k-1)\pi}{2}} = \begin{cases} +\dfrac {k(k-1)}{2} & \text{if } \dfrac {k(k-1)}{2} \text{ is even} \\ -\dfrac {k(k-1)}{2} & \text{if } \dfrac {k(k-1)}{2} \text{ is odd} \end{cases}

∑ k = 19 98 A k = − 19 ( 18 ) 2 + 20 ( 19 ) 2 + 21 ( 20 ) 2 − 22 ( 21 ) 2 − . . . − 98 ( 97 ) 2 = 19 − 21 + 23 − 25 + 27 − . . . − 93 + 95 − 97 = − 2 − 2 − 2 − . . . − 2 = 20 ( − 2 ) = − 40 \begin{aligned} \displaystyle \sum_{k=19}^{98} {A_k} & = -\frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - ... - \frac{98(97)}{2} \\ & = 19 -21 + 23 - 25 + 27 -... -93+95-97 \\ & = - 2 -2 - 2 -... -2 = 20(-2) = -40 \end{aligned}

⇒ ∣ A 19 + A 20 + A 21 + . . . + A 98 ∣ = ∣ − 40 ∣ = 40 \Rightarrow |A_{19}+A_{20}+A_{21}+...+A_{98}| = |-40| = \boxed{40}

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ctr = 0
for i in range(19,99):
    n = i*(i-1)/2
    if n%2:
        ctr -= n
    else:
        ctr += n

print ctr

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Refaat M. Sayed - 6 years, 1 month ago

A calculator.

Edwin Ma - 6 years, 1 month ago

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