Will you integrate for each element?

Calculus Level 4

a k = ∫ 0 π 2 ( k + sin ⁡ θ ) 2 cos ⁡ θ d θ \large a_k = \displaystyle \int_{0}^{\frac{\pi}{2}} (k + \sin\theta)^2\cos\theta \, d\theta

For a k a_k , where k k is a positive integer, as defined above, then the determinant

∣ a k k k 2 a k + 1 k + 1 ( k + 1 ) 2 a k + 2 k + 2 ( k + 2 ) 2 ∣ = p q \begin{vmatrix} a_k & k & k^2 \\ a_{k+1} & k+1 & (k+1)^2 \\ a_{k+2} & k+2 & (k+2)^2 \end{vmatrix} = \frac pq

where p p and q q are coprime positive integers . Find p + q p + q .

Try my set .


The answer is 5.

Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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1 solution

Chew-Seong Cheong
Jul 31, 2016

a k = ∫ 0 π 2 ( k + sin ⁡ θ ) 2 cos ⁡ θ d θ Let x = sin ⁡ θ ⟹ d x = cos ⁡ θ d θ = ∫ 0 1 ( k + x ) 2 d x = ∫ 0 1 ( k 2 + 2 k x + x 2 ) d x = k 2 x + k x 2 + x 3 3 ∣ 0 1 = k 2 + k + 1 3 \begin{aligned} a_k & = \int_0 ^\frac \pi 2 (k + \sin \theta)^2 \cos \theta \ d \theta & \small \color{#3D99F6}{\text{Let }x = \sin \theta \implies dx = \cos \theta \ d \theta} \\ & = \int_0 ^1 (k+x)^2 \ dx \\ & = \int_0 ^1 (k^2+2kx + x^2) \ dx \\ & = k^2x+kx^2 + \frac {x^3}3 \bigg|_0^1 \\ & = k^2+k + \frac 13 \end{aligned}

X = ∣ a k k k 2 a k + 1 k + 1 ( k + 1 ) 2 a k + 2 k + 2 ( k + 2 ) 2 ∣ = ∣ k 2 + k + 1 3 k k 2 ( k + 1 ) 2 + k + 1 + 1 3 k + 1 ( k + 1 ) 2 ( k + 2 ) 2 + k + 2 + 1 3 k + 2 ( k + 2 ) 2 ∣ { r o w 1 → r o w 1 r o w 2 − r o w 1 → r o w 2 r o w 3 − r o w 2 → r o w 3 = ∣ k 2 + k + 1 3 k k 2 2 k + 2 1 2 k + 1 2 k + 4 1 2 k + 3 ∣ { r o w 1 → r o w 1 r o w 2 → r o w 2 r o w 3 − r o w 2 → r o w 3 = ∣ k 2 + k + 1 3 k k 2 2 k + 2 1 2 k + 1 2 0 2 ∣ = 2 k 2 + 2 k + 2 3 + 4 k 2 + 2 k − 4 k 2 − 4 k − 2 k 2 = 2 3 \begin{aligned} X & = \begin{vmatrix} a_k & k & k^2 \\ a_{k+1} & k+1 & (k+1)^2 \\ a_{k+2} & k+2 & (k+2)^2 \end{vmatrix} \\ & = \begin{vmatrix} k^2+k + \frac 13 & k & k^2 \\ (k+1)^2+k+1 + \frac 13 & k+1 & (k+1)^2 \\ (k+2)^2+k + 2 + \frac 13 & k+2 & (k+2)^2 \end{vmatrix} & \small \color{#3D99F6}{\begin{cases} row_1 \to row_1 \\ row_2-row_1 \to row_2 \\ row_3-row_2 \to row_3 \end{cases}} \\ & = \begin{vmatrix} k^2+k + \frac 13 & k & k^2 \\ 2k+2 & 1 & 2k+1 \\ 2k+4 & 1 & 2k+3 \end{vmatrix} & \small \color{#3D99F6}{\begin{cases} row_1 \to row_1 \\ row_2 \to row_2 \\ row_3-row_2 \to row_3 \end{cases}} \\ & = \begin{vmatrix} k^2+k + \frac 13 & k & k^2 \\ 2k+2 & 1 & 2k+1 \\ 2 & 0 & 2 \end{vmatrix} \\ & = 2k^2+2k + \frac 23 + 4k^2+2k - 4k^2- 4k - 2k^2 \\ & = \frac 23 \end{aligned}

⟹ p + q = 2 + 3 = 5 \implies p+q = 2+3 = \boxed{5}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

You could put k=0 as the answer is independent of k. But that would be 'cheating'.

Ajinkya Shivashankar - 4 years, 7 months ago

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There may be a easier way to solve the determinant. I am no good with it.

Chew-Seong Cheong - 4 years, 7 months ago

Not exactly k = 0 k=0 since the question states that k k is a positive integer and not "non negative integer". But we could substitute k = 1 k=1 , Cheers!!! :-)

Rohan Shinde - 2 years, 5 months ago

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