Let - a-k-k-10C-k- b-k-10 -k-10C-k-and - A-k- a-k amp-0- 0 amp-b-k -If -160- - A-k-1-9A-k- a amp-0- 0 amp-b - find the value of - a - b-

a + b =∑_k=1^9 (a_k+b_k) #160; #160; #160; #160; #160; #160; #160; =10∑_k=1^9 (^10C_k)=10(2^10 2) #160; #160; #160; #160; #1 ...

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Multiplication of Matrices

Let a_k=...

Question

Let $a_{k} = k (^{10} C_{k}), b_{k} = (10 - k) (^{10} C_{k})$
and $A_{k} = [\begin{matrix} a_{k} & 0 0 & b_{k} \end{matrix}]$
If $A = 9 \sum k = 1 A_{k} = [\begin{matrix} a & 0 0 & b \end{matrix}]$ , find the value of $a + b .$

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a_{k} =^{n} C_{k}

0 \leq k \leq n

A_{k} = [\begin{matrix} a_{k - 1} & 0 0 & a_{k} \end{matrix}]

1 \leq k \leq n

B = n - 1 \sum k = 1 A_{k} . A_{k + 1} = [\begin{matrix} a & 0 0 & b \end{matrix}]

a > b, k > 0 \Rightarrow {\begin{matrix} a k > b k - a k < - b k \end{matrix}

k > 0

a \geq b \Rightarrow \frac{a}{k} \geq \frac{b}{k}

k < 0

\frac{sin x}{a} = \frac{cos x}{b} = \frac{tan x}{c} = k

b c + \frac{1}{c k} + \frac{a k}{1 + b k}

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