1- 2 -1 2- 3 - 3-2- 2 12 - 5- 2 -7 24- 3 - 17-12- 2 144 -

(√( 2 ) 1)^ 2 =3 2√( 2 ) , (√( 2 ) 1)^ 2 =5√( 2 ) 7 (√( 2 ) 1)^ 4 =(3 2√( 2 ) )^ 3 =17 12√( 2 ) Also (2√( 3 ) )^ 2 =12 , (2√( 3 ) ) ...

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Byju's Answer

Standard XII

Mathematics

General Tips for Choosing Function

1+√ 2 -1 2√ 3...

Question

$1 + \frac{\sqrt{2} - 1}{2 \sqrt (3)} + \frac{3 - 2 \sqrt{2}}{12} + \frac{5 \sqrt{2} - 7}{24 \sqrt (3)} + \frac{17 - 12 \sqrt{2}}{144} + . . . . \infty$

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Solution

$(\sqrt{2} - 1)^{2} = 3 - 2 \sqrt{2}$ , $(\sqrt{2} - 1)^{2} = 5 \sqrt{2} - 7$
$(\sqrt{2} - 1)^{4} = (3 - 2 \sqrt{2})^{3} = 17 - 12 \sqrt{2}$
Also $(2 \sqrt{3})^{2} = 12$ , $(2 \sqrt{3})^{3} = 24 \sqrt{3}$
$(2 \sqrt{3})^{4} = 144$ etc.
Hence the given series is a G.P. of infinite number terms whose first term is $1$ and $r = \frac{\sqrt{2} - 1}{2 \sqrt (3)} < 1$
$∴ S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{\sqrt (2) - 1}{2 \sqrt (3)}} = \frac{2 \sqrt (3)}{2 \sqrt (3) - \sqrt (2) + 1}$

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Q. Find the sum of the following infinite series

1 + \frac{\sqrt{2} - 1}{2 \sqrt{3}} + \frac{3 - 2 \sqrt{2}}{12} + \frac{5 \sqrt{2} - 7}{24 \sqrt{3}} + \frac{17 - 12 \sqrt{2}}{144} + . . . . .

$If m = \frac{1}{3 - 2 \sqrt{2}} then, m^{2} =$ _______.

The sum of the infinite series

$[1 + \frac{\sqrt{2} - 1}{2 \sqrt{2}} + \frac{3 - 2 \sqrt{2}}{12} + \frac{5 \sqrt{2} - 7}{24 \sqrt{2}} + \frac{17 - 2 \sqrt{2}}{80} + . . . \infty]$ is $a - \frac{1}{b} (\sqrt{c} + 1) {log}_{e} 2$

Find $(a + b)^{2} + c^{2}$

Evaluate :

$(i) \frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9} (i i) \frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5} (i i i) \frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9} (i v) \frac{3}{8} + \frac{- 5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{- 5}{8} + \frac{- 2}{7}$

$12 \times (2 - \frac{1}{3}) \times (2 - \frac{3}{5}) \times (2 - \frac{5}{7}) \times (2 - \frac{7}{9}) \times (2 - \frac{9}{11}) \dots \dots \dots (2 - \frac{997}{999}) \times (\frac{1}{2002})$ = ?

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1+√2−12√(3)+3−2√212+5√2−724√(3)+17−12√2144+....∞

(√2−1)2=3−2√2, (√2−1)2=5√2−7(√2−1)4=(3−2√2)3=17−12√2Also (2√3)2=12, (2√3)3=24√3(2√3)4=144 etc.Hence the given series is a G.P. of infinite number terms whose first term is 1 and r=√2−12√(3)<1∴S∞=a1−r=11−√(2)−12√(3)=2√(3)2√(3)−√(2)+1

$1 + \frac{\sqrt{2} - 1}{2 \sqrt (3)} + \frac{3 - 2 \sqrt{2}}{12} + \frac{5 \sqrt{2} - 7}{24 \sqrt (3)} + \frac{17 - 12 \sqrt{2}}{144} + . . . . \infty$