GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 2

Answer 1

Comments

how it comes I cant understand plz if possible then give me detail solution. I unable to understand steps in between

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See follow steps:-

1) Take the sequence as mentioned in question.( this Sequence is in the AGP form)....(eq1)

2) find out the ratio of that sequence. Its  r 

3) take another equation(eq2).     eq2= r(eq1)... okay

4) now substract eq1 - eq2. 

5) we get another equation which is in the GP Form. okay...

6) now we know how to find sum in GP apply this formula  a/(1-r) ( as its in infinite terms).

   in GP a= first value its 1

             r = br

now apply this all steps according you will understand!!

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@Skb55871 in step 6 the formula would be a/(1-r)

Answer 2

$1 + (1 + b)r + (1 + b + b^2)r^2 + (1 + b + b^2 + b^3)r^3 + ... \infty$

= $\sum_{i=0}^{\infty} (\sum_{j=0}^{i} b^j)r^i$

= $\sum_{i=0}^{\infty} (\frac{1-b^{i+1}}{1-b})r^i$

= $\frac{1}{1-b} \sum_{i=0}^{\infty} (r^i - b^{i+1}r^i)$

= $\frac{1}{1-b} ( \sum_{i=0}^{\infty} r^i - b\sum_{i=0}^{\infty} b^i r^i)$

= $\frac{1}{1-b} \times (\frac{1}{1-r} - \frac{b}{1-br})$

= $\frac{1}{1-b} \times \frac{1-b}{(1-r)(1-br)}$

= $\frac{1}{(1-r)(1-br)}$

Comments

This is really intuitive, but this is not correct.

We only know that '|br|<1' , so we can directly simplify the series as '1/1-br' in the subsequent step, but We don't know anything about 'r' and 'b' , and whether '|r|<1' or not. So we cannot simplify the series as '1/1-r' in the subsequent step.

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If $ |r| \ge 1 $, then the given series will never converge. So, $|r|$ must be  $< 1$.

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bro but he solved using gp. in gp it doesnt matter if it less than 1 or not as it is infinity series.

Answer 3

Let     ${S=1 + (1+b)r + (1+b+b}$$^{2}$${)r}$$^{2}$${+…..}$

 

${S=1 + r+br + r}$$^{2}$${+br}$$^{2}$${+b}$$^{2}$${r}$$^{2}$${+r}$$^{3}$${+br}$$^{3}$${+b}$$^{2}$${r}$$^{3}$${+b}$$^{3}$${r}$$^{3}$${+…..}$

${S=1 + r+r}$$^{2}$${+……..+br(1+}$${r+}$${r}$$^{2}$${+…..)+b}$$^{2}$${r}$$^{2}$${(1+}$${r+}$${r}$$^{2}$${+…..)+b}$$^{3}$${r}$$^{3}$${(1+r+r}$$^{2}$${+ …..)+…..}$

${S=(1+r+r}$$^{2}$${+…...)(1+br+b}$$^{2}$${r}$$^{2}$${+b}$$^{3}$${r}$$^{3}$${+…...)}$

now let ${X= 1+r+r}$$^{2}$${+….}$

then ${Xr=r+r}$$^{2}$${+r}$$^{3}$${+….}$

${X-Xr=1}$

${X = 1/(1-r)}$

 

Now let ${Y=1+br+b}$$^{2}$${r}$$^{2}$${+….}$

then ${Ybr=br + b}$$^{2}$${r}$$^{2}$${+b}$$^{3}$${r}$$^{3}$${+….}$

${Y-Ybr=1}$

${Y=1/(1-br)}$

 

Since ${S= XY}$

${S=1/(1-r)(1-br)}$

${Ans : A}$

Comments

I solved like this only, but this method is not correct.

We only know that '|br|<1' , so we can directly simplify the series as '1/1-br' in the subsequent step, but We don't know anything about 'r' and 'b' , and whether '|r|<1' or not. So we cannot simplify the series as '1/1-r' in the subsequent step.