∑_ k=1 ^ 40 ( 1+(1)/(k^2)+(1)/((k+1)^2) )=a+(b)/(c) where a, b, c ∈ ℕ, b < c, gcd(b, c)=1 , then what is the value of a+b ?

Correct answer: 80 Let aligned t_k & = 1+(1)/(k^2)+(1)/((k+1)^2) & = (k^2(k+1)^2+k^2+(k+1)^2)/(k^2(k+1)^2) & = (k^4+2 k^3+3 k^2+2 k+1)/(k^2(k+1)^2) = ((k^2+k+1)^2)/(k^2(k+1)^2) & =(k^2+k+1)/(k(k+1))=1+(1)/(k(k+1)) & =1+(1)/(k)-(1)/(k+1) aligned aligned & S_n=∑_ k=1 ^n t_k=t_1+t_2+ +t_n & =n+1-(1)/(n+1)=(n+1)-(1)/(n+1)=n+(n)/(n+1) & ∑_ k=1 ^ 40 1+(1)/(k^2)+(1)/((k+1)^2) =40+(40)/(41) & a=b=40, c=41 & a+b=80 . aligned

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

∑_ k=1 ^ 40 ( 1+(1)/(k^2)+(1)/((k+1)^2) )=a+(b)/(c) where a, b, c ∈...

\sum_{k=1}^{40}

\left(\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}\right)=a+\frac{b}{c}

where

a, b, c \in \mathbb{N}, b < c, \operatorname{gcd}(b, c)=1

, then what is the value of

a+b

Correct answer

Step-by-step solution

Let

\begin{aligned} t_k & =\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}} \\ & =\sqrt{\frac{k^2(k+1)^2+k^2+(k+1)^2}{k^2(k+1)^2}} \\ & =\sqrt{\frac{k^4+2 k^3+3 k^2+2 k+1}{k^2(k+1)^2}}=\sqrt{\frac{\left(k^2+k+1\right)^2}{k^2(k+1)^2}} \\ & =\frac{k^2+k+1}{k(k+1)}=1+\frac{1}{k(k+1)} \\ & =1+\frac{1}{k}-\frac{1}{k+1} \end{aligned}

\begin{aligned} & \therefore \quad S_n=\sum_{k=1}^n t_k=t_1+t_2+\ldots \ldots+t_n \\ & \quad=n+1-\frac{1}{n+1}=(n+1)-\frac{1}{n+1}=n+\frac{n}{n+1} \\ & \quad \sum_{k=1}^{40} \sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}=40+\frac{40}{41} \\ & \therefore \quad a=b=40, c=41 \\ & \therefore \quad a+b=80 .\end{aligned}

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