Algebra (Olympiad)hardFree

Telescoping Nested Square Root Sum: Find 100(x-[x]) | IOQM

JEE Maths question with a full step-by-step solution.

Question
Let
x=1+112+122+1+122+132+1+132+142+⋯+1+1992+11002.x=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\cdots+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}.
 If [x][x] denotes the greatest integer not exceeding xx, then find the value of 100 (x−[x])100\,(x-[x])
Solution
Answer: 99
Step 1: Simplify the general term 1+1n2+1(n+1)2\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}} by writing the expression under the root as a perfect square. Since 1n−1n+1=1n(n+1)\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{n(n+1)},
1n2+1(n+1)2=(1n−1n+1)2+2n(n+1)=(1n(n+1))2+2n(n+1).\frac{1}{n^2}+\frac{1}{(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2+\frac{2}{n(n+1)}=\left(\frac{1}{n(n+1)}\right)^2+\frac{2}{n(n+1)}.
 Step 2: Add the leading 11 to complete the square:
1+1n2+1(n+1)2=1+2⋅1n(n+1)+(1n(n+1))2=(1+1n(n+1))2.1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=1+2\cdot\frac{1}{n(n+1)}+\left(\frac{1}{n(n+1)}\right)^2=\left(1+\frac{1}{n(n+1)}\right)^2.
 Step 3: Take the positive square root and split the fraction using 1n(n+1)=1n−1n+1\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}:
1+1n2+1(n+1)2=1+1n(n+1)=1+1n−1n+1.\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1}.
 Step 4: Sum from n=1n=1 to n=99n=99. The constant 11 contributes 9999, and the remaining part telescopes:
x=∑n=199(1+1n−1n+1)=99+[(11−12)+(12−13)+⋯+(199−1100)].x=\sum_{n=1}^{99}\left(1+\frac{1}{n}-\frac{1}{n+1}\right)=99+\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{99}-\frac{1}{100}\right)\right].
 Step 5: Only the first and last fractions survive the cancellation:
x=99+(11−1100)=99+99100.x=99+\left(\frac{1}{1}-\frac{1}{100}\right)=99+\frac{99}{100}.
 Step 6: Since 99<x=99+99100<10099<x=99+\dfrac{99}{100}<100, the greatest integer is [x]=99[x]=99, so
x−[x]=99100.x-[x]=\frac{99}{100}.
 Step 7: Therefore
100 (x−[x])=100×99100=99.100\,(x-[x])=100\times\frac{99}{100}=99.
 Answer: 99
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