Question #8e330

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Question #8e330

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Answer 1 · 2017-05-28T14:05:15

$2.$

Explanation:

The Expression
$=1/(sqrt2+sqrt1)+1/(sqrt3+sqrt2)+1/(sqrt4+sqrt3)+...+1/(sqrt9+sqrt8),$

$=1/(sqrt2+sqrt1)xx(sqrt2-sqrt1)/(sqrt2-sqrt1)$
$+1/(sqrt3+sqrt2)xx(sqrt3-sqrt2)/(sqrt3-sqrt2)$
$+1/(sqrt4+sqrt3)xx(sqrt4-sqrt3)/(sqrt4-sqrt3)$
$+vdots$
$+1/(sqrt9+sqrt8)xx(sqrt9-sqrt8)/(sqrt9-sqrt8)$

$=(sqrt2-sqrt1)/(2-1)+(sqrt3-sqrt2)/(3-2)+...+(sqrt9-sqrt8)/(9-8)$

$=(cancelsqrt2-1)+(cancelsqrt3-cancelsqrt2)+(cancelsqrt4-cancelsqrt3)+...+(sqrt9-cancelsqrt8)$

$=3-1$

$=2.$

Enjoy Maths.!

Answer 2 · 2017-05-28T14:05:44

The answer is $=2$

Explanation:

We need

$(a+b)(a-b)=a^2-b^2$

The general term is

$u_n=1/(sqrtn+sqrt(n+1))$

Simplifying the denominator

$u_n=1/(sqrt(n+1)+sqrt(n))*(sqrt(n+1)-sqrtn)/(sqrt(n+1)-sqrtn)$

$=(sqrt(n+1)-sqrtn)/(n+1-n)$

$=(sqrt(n+1)-sqrtn)$

Therefore,

$n=1$ , $=>$ , $u_1=cancelsqrt2-1$

$n=2$ , $=>$ , $u_2=cancelsqrt3-cancelsqrt2$

$n=3$ , $=>$ , $u_3=cancelsqrt4-cancelsqrt3$

$n=7$ , $=>$ , $u_7=cancelsqrt8-cancelsqrt7$

$n=8$ , $=>$ , $u_8=sqrt9-cancelsqrt8$

$sum_(n=1)^8u_n=sqrt9-1=3-1=2$

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Question #8e330

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