Sequences And Series #2

Algebra Level 3

If a n = ∑ k = 1 n 1 + 1 k 2 + 1 ( k + 1 ) 2 a_{n}=\displaystyle\sum_{k=1}^n \sqrt{1+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}} find 6 a 5 6a_{5}


The answer is 35.

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1 solution

First , lets make a substitution to simplify the expression :

k = w − 1 2 k=w-\dfrac{1}{2}

The expression now becomes :

( w 2 − 1 4 ) 2 + ( w − 1 2 ) 2 + ( w + 1 2 ) 2 ( w 2 − 1 4 ) 2 \sqrt{\dfrac{(w^2-\frac{1}{4})^2+ (w-\frac{1}{2})^2+(w+\frac{1}{2})^2}{(w^2-\frac{1}{4})^2}}

Simplifying , we get :

( w 2 + 3 4 ) 2 ( w 2 − 1 4 ) 2 \sqrt{\dfrac{(w^2+\frac{3}{4})^2}{(w^2-\frac{1}{4})^2}}

w 2 + 3 4 w 2 − 1 4 = 1 + 1 ( w − 1 2 ) ( w + 1 2 ) = 1 + 1 k ( k + 1 ) \dfrac{w^2+\frac{3}{4}}{w^2-\frac{1}{4}} = 1+\dfrac{1}{(w-\frac{1}{2})(w+\frac{1}{2})}=1+\dfrac{1}{k(k+1)}

Now, it is simple telescoping and we get :

a n = n ( n + 2 ) n + 1 \boxed{a_{n}=\dfrac{n(n+2)}{n+1}}

So, 6 a 5 = 35 \boxed{6a_{5}=35} .

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