If Ak- k k-1- k-1 k - then -A1-A2-A2015-

The correct option is B (2015)^2 [ [ A_ 1 ] =| [ [ 1 amp; 0; 0 amp; 1 ]] #160; | =1; ...

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Standard XII

Mathematics

Summation by Sigma Method

If Ak=[ k ...

Question

If $A_{k} = [\begin{matrix} k & k - 1 k - 1 & k \end{matrix}]$ then $| A_{1} | + | A_{2} | + . . + | A_{2015} | = ?$

0

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2015

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(2015)^{2}

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(2015)^{3}

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Solution

The correct option is B $(2015)^{2}$
$\begin{matrix} [A_{1}] = ∣ ∣ ∣ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] ∣ ∣ ∣ = 1 [A_{2}] = ∣ ∣ ∣ [\begin{matrix} 2 & 1 1 & 2 \end{matrix}] ∣ ∣ ∣ = 3 [A_{3}] = ∣ ∣ ∣ [\begin{matrix} 3 & 2 2 & 3 \end{matrix}] ∣ ∣ ∣ = 5 A_{2015} = ∣ ∣ ∣ [\begin{matrix} 2015 & 2014 2014 & 2015 \end{matrix}] ∣ ∣ ∣ = {(2015)}^{2} - {(2014)}^{2} = (4029) ∴ | A_{1} | + | A_{3} | + . . . . . . . | A_{2015} | 1 + 3 + 5 + 7....4029 T h e s e s e r i e s a r e t e r m = 1 c o m m o n d i f f e r e n c e = 2 \Rightarrow 4029 = 1 + (n - 1) \times 2, \frac{4028}{2} = n - 1, n = 2015 \Rightarrow S u m = \frac{2015}{2} (1 + 4029) \Rightarrow \frac{2015 \times 4030}{2} = (2015)^{2} \end{matrix}$

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