The value of 1/√2 +√1 + 1/√2+√3 + 1/√3+√4+....1/√8+√9 is equal to (a)5/√2 (b)5/√8 (c)2 (d)4 ??

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Answer 1 · 2018-05-05T07:38:28

The Right Option is (c) #2.#

Note that, #AA n in NN, 1/(sqrt(n+1)+sqrtn)#,

#=1/(sqrt(n+1)+sqrtn)xx{(sqrt(n+1)-sqrtn)}/{(sqrt(n+1)-sqrtn)}#,

#={(sqrt(n+1)-sqrtn)}/{(n+1)-n}#.

Thus, #1/(sqrtn+sqrt(n+1))=sqrt(n+1)-sqrtn ; (n in NN)......(ast)#.

Using #(ast)" for "n=1,2,...,8#, we have,

#1/(sqrt1+sqrt2)+1/(sqrt2+sqrt3)+1/(sqrt3+sqrt4)+...+1/(sqrt8+sqrt9)#,

#=(cancelsqrt2-sqrt1)+(cancelsqrt3-cancelsqrt2)+(cancelsqrt4-cancelsqrt3)+...+(sqrt9-cancelsqrt8)#

#=sqrt9-sqrt1#,

#=3-1#,

#2#.

So, the Right Option is (c) #2.#