1-3/16+1·4/1·2≤ft(3/16)²-1·4·7/1·2·3≤ft(3/16)³+·s =

Correct Answer is : (45)2/3≤ft(4/5)²/3}(54)2/3

Correct Answer is : (45)2/3≤ft(4/5)²/3}(54)2/3

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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

1-\frac{3}{16}+\frac{1\cdot4}{1\cdot2}\left(\frac{3}{16}\right)^2-\frac{1\cdot4\cdot7}{1\cdot2\cdot3}\left(\frac{3}{16}\right)^3+\cdots =

Solution:

We have,

1-\;\frac{3}{16}+\;\frac{1 \cdot 4}{1 \cdot 2}\left(\;\frac{3}{16}\right)^{2}-\;\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\;\frac{3}{16}\right)^{3}+\dots =

We know that,

(1-x)^{n}=1-n x+\;\frac{n(n-1)}{2!}x^{2}-\;\frac{n(n-1)(n-2)}{3!}x^{3}

Let

(1-x)^{n}=1-\;\frac{3}{16}+\;\frac{1\cdot4}{1\cdot2}\left(\;\frac{3}{16}\right)^{2}-\;\frac{1\cdot4\cdot7}{1\cdot2\cdot3}\left(\;\frac{3}{16}\right)^{3}\cdots

\therefore \;\; n x=\;\frac{3}{16}

and

\;\frac{n(n-1)}{2!}x^{2}=\;\frac{1\cdot4}{2!}\left(\;\frac{3}{16}\right)^{2}

n x=\;\frac{3}{16}

and

n(n-1)x^{2}=4\left(\;\frac{3}{16}\right)^{2}

n^{2}\cdot x^{2}=\;\frac{3^{2}}{16}

and

n(n-1)x^{2}=4\left(\;\frac{3}{16}\right)^{2}

n(n-1)x^{2}=4n^{2}x^{2}

n^{2}-n=4n^{2}

n-1=4n \Rightarrow n=-\;\frac{1}{3}

x=\;\frac{3}{16}\times-3=\;\frac{-9}{16}

(1-x)^{n}=\left(1+\;\frac{9}{16}\right)^{-1/3}=\left(\;\frac{25}{16}\right)^{-1/3}=\left(\;\frac{4}{5}\right)^{2/3}

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