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Let $|z| < 1$. Define $M_{n}(z)= \sum_{i=1}^{10} z^{10^{n}(i - 1)}?$ what is

             $\prod_{i=0}^{\infty} M_{i}(z)= M_{0}(z)\times M_{1}(z) \times M_{2}(z) \times ...?$

  1. Can't be determined.
  2. $1/ (1 - z)$
  3. $1/ (1 + z)$
  4. $1 - z^{9}$
  5. None of the above.
in Numerical Ability
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2 Answers

6 votes
 
Best answer
$M_{n}(z) = \sum_{i=1}^{10}z^{10^{n}(i-1)}$

$\quad \quad =z^{0*10^{n}}+z^{1*10^{n}}+z^{2*10^{n}}+\ldots +z^{9*10^{n}}$

$\quad \quad =1+z^{1*10^{n}}+z^{2*10^{n}}+\ldots +z^{9*10^{n}}$

$\quad\quad = \dfrac{1-(z^{10^{n}})^{10}}{1-z^{10^{n}}}$

$\quad \quad = \dfrac{1-z^{10^{(n+1)}}}{1-z^{10^{n}}}$  
$M_{n}(z)$ $= \dfrac{1-z^{10^{(n+1)}}}{1-z^{10^{n}}}$  
$Now, $

$\prod_{i=0}^{\infty}M_{i}(z)= M_{0}(z){\times}M_{1}(z){\times}M_{2}(z){\times}\ldots $

$\quad = \left(\dfrac{1-z^{10^{1}}}{1-z^{10^{0}}}\right){\times}\left(\dfrac{1-z^{10^{2}}}{1-z^{10^{1}}} \right ){\times}\left(\dfrac{1-z^{10^{3}}}{1-z^{10^{2}}} \right ){\times}\ldots {\times}\left(\dfrac{1-z^{10^{k}}}{1-z^{10^{(k-1)}}} \right ){\times}\left(\dfrac{1-z^{10^{(k+1)}}}{1-z^{10^{k}}} \right ){\times}\ldots$

$\quad = \dfrac{1}{1-z}$

for ending terms, as $ |z|<1, z^{\infty}$ tends to $0,$ $1-z^{\infty} $ tends to $1.$

Correct Answer: $B$

edited by
0

@ankitgupta.1729

Which formula is used in the $\mathbf{3^{rd}}$ line?

0

@ankitgupta.1729

Which formula is used in the $\mathbf{4^{th}}$ line?

2

@`JEET sum of 10 terms of GP, with first term = 1 and common ratio = $z^{10^n}$

1
Yes, thanks!

I was applying $\mathbf{1+x+x^2+\dots}$
1 vote
Answer:

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