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Question Number 9460 by tawakalitu last updated on 09/Dec/16

	Answered by sou1618 last updated on 09/Dec/16
	$(√(11−2))=(√(10+1−2))=(√9)=3 (√(1111−22))=(√(1000+100+10+1−20−2))=(√(1089))=33 ...... X=(√(1111......11_(2000digits) −22...2_(1000digits) )) ∗1000_(4digits) =10^(4−1) X^2 =(10^(1999) +10^(1998) +...+10^1 +10^0 )−2(10^(999) +10^(998) +...+10^1 +10^0 ) =Σ_(k=1) ^(2000) 10^(k−1) −2Σ_(k=1) ^(1000) 10^(k−1) =((10^(2000) −1)/(10−1))−2((10^(1000) −1)/(10−1)) =(1/9)(10^(2000) −2×10^(1000) +1) =(1/3^2 ){(10^(1000) )^2 −2(10^(1000) )+1} =(1/3^2 )(10^(1000) −1)^2 ={(1/3)(1000...000_(1001digits) −1)}^2 X=(1/3)×999...999_(1000digits) =333...333_(1000digits) (√(11...1_(2n(digits)) −22...2_(n(digits)) ))=33...3_(n(digits))$
	$\sqrt{11 - 2} = \sqrt{10 + 1 - 2} = \sqrt{9} = 3$ $\sqrt{1111 - 22} = \sqrt{1000 + 100 + 10 + 1 - 20 - 2} = \sqrt{1089} = 33$ $. . . . . .$ $X = \sqrt{1111 . . . . . . 11_{2000 d i g i t s} - 22 . . . 2_{1000 d i g i t s}}$ $* 1000_{4 d i g i t s} = 10^{4 - 1}$ $X^{2} = (10^{1999} + 10^{1998} + . . . + 10^{1} + 10^{0}) - 2 (10^{999} + 10^{998} + . . . + 10^{1} + 10^{0})$ $= \underset{k = 1}{\sum^{2000}} 10^{k - 1} - 2 \underset{k = 1}{\sum^{1000}} 10^{k - 1}$ $= \frac{10^{2000} - 1}{10 - 1} - 2 \frac{10^{1000} - 1}{10 - 1}$ $= \frac{1}{9} (10^{2000} - 2 \times 10^{1000} + 1)$ $= \frac{1}{3^{2}} {{(10^{1000})}^{2} - 2 (10^{1000}) + 1}$ $= \frac{1}{3^{2}} {(10^{1000} - 1)}^{2}$ $= {\frac{1}{3} (1000 . . . 000_{1001 d i g i t s} - 1)}^{2}$ $X = \frac{1}{3} \times 999 . . . 999_{1000 d i g i t s}$ $= 333 . . . 333_{1000 d i g i t s}$ $\sqrt{11 . . . 1_{2 n (d i g i t s)} - 22 . . . 2_{n (d i g i t s)}} = 33 . . . 3_{n (d i g i t s)}$
	Commented by tawakalitu last updated on 09/Dec/16

	$Thank you sir . God bless you .$
	Commented by tawakalitu last updated on 10/Dec/16

	$i really appreciate .$
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