Can you Identify?

Calculus Level 4

∑ n = 1 49 1 n + ( n − 1 ) ( n + 1 ) = a + b c \sum_{n = 1}^{49} \frac{1}{\sqrt{n + \sqrt{(n - 1)(n + 1)}}} = a + b\sqrt{c}

a , b , c a, b, c are positive integers, and c c is not divisible by the square of any prime.

find [ a + b + c − 2.123456789 ] \Big[\frac{a+ b + c}{-2.123456789}\Big]

[.] denotes greatest integer function

Please Post Your Solutions


The answer is -5.

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U Z by U Z , 24, Guyana

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1 solution

U Z
Oct 28, 2014

T n = 1 n + ( n 2 − 1 ) T_{n} = \dfrac{1}{\sqrt{n + \sqrt{(n^{2} - 1)}}}

= 1 ( n + 1 2 + n − 1 2 ) 2 = \dfrac{1}{\sqrt{\left(\sqrt{\dfrac{n+1}{2}} + \sqrt{\dfrac{n -1}{2}}\right)^{2}}}

= 1 n + 1 2 + n − 1 2 =\dfrac{1}{\sqrt{\dfrac{n + 1}{2}} + \sqrt{\dfrac{n - 1}{2}}}

= n + 1 2 − n − 1 2 n + 1 2 − n − 1 2 = \dfrac{\sqrt{\dfrac{n + 1}{2}}- \sqrt{\dfrac{n - 1}{2}}}{\dfrac{n + 1}{2} - \dfrac{n - 1}{2}}

= n + 1 2 − n − 1 2 = \sqrt{\dfrac{n + 1}{2}} - \sqrt{\dfrac{n - 1}{2}}

= 2 2 − 0 + 3 2 − 1 2 + . . . . . . . . . . . + 50 2 − 48 2 = \sqrt{\dfrac{2 }{2}} - 0 + \sqrt{\dfrac{3 }{2}} - \sqrt{\dfrac{1 }{2}} + ...........+ \sqrt{\dfrac{50 }{2}} - \sqrt{\dfrac{48 }{2}}

= 50 2 + 49 2 − 1 2 − 0 = 5 + 3 2 = \sqrt{\dfrac{50 }{2}} + \sqrt{\dfrac{49 }{2}} - \sqrt{\dfrac{1 }{2}} - 0 = 5 + 3\sqrt{2}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Why is it tagged in calculus?

Kartik Sharma - 6 years, 7 months ago

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Sequence and series comes under calculus part( here on brilliant.org) ,

but I too agree because in India it is taught under algebra

U Z - 6 years, 7 months ago

well there is a very useful formula too which I want to share. It says -

a + b c = a + b 2 − a 2 + a − b 2 − a 2 \sqrt{a + b\sqrt{c}} = \sqrt{\frac{a + \sqrt{{b}^{2} - a}}{2}} + \sqrt{\frac{a - \sqrt{{b}^{2} - a}}{2}}

Kartik Sharma - 6 years, 7 months ago

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If in some question we get the RHS as some function or equation and we want to reduce it to a simpler form like that in the LHS , then what will be the value of c with respect to a and b

Vighnesh Raut - 6 years, 5 months ago

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