\[ \large a = 1 + \dfrac1{2^2} + \dfrac1{3^2} + \cdots + \dfrac1{2016^2} \]

Find the value of \(\lfloor a \rfloor \).

**Notation**: \( \lfloor \cdot \rfloor \) denotes the floor function.

\[ \large a = 1 + \dfrac1{2^2} + \dfrac1{3^2} + \cdots + \dfrac1{2016^2} \]

Find the value of \(\lfloor a \rfloor \).

**Notation**: \( \lfloor \cdot \rfloor \) denotes the floor function.

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TopNewestThe answer is 1.

Note that if \[s_n=\sum_{r=1}^n \frac{1}{r^2} \quad \forall n \in \mathbb{N}-\{1\} \\ s_1=1\] then \(\{s_n\}_{n=1}^{\infty}\) is a strictly increasing sequence that which is always less than 2, that is, \[n<m \implies s_n<s_m \quad n, m \in \mathbb{N} \\ s_n<2 \quad \forall n \in \mathbb{N}\]

Here are the proofs:

The first statement is true as \(s_m\) contains more number of positive terms than \(s_n\).

To prove the next assertion, note that \[\begin{align}0&<r(r-1)<r^2 \quad \forall r \in \mathbb{N}-\{1\} \\\implies \frac{1}{r^2} &< \frac{1}{r(r-1)} \quad \forall r \in \mathbb{N}-\{1\} \\\implies s_n&<1+\sum_{r=2}^n \frac{1}{r(r-1)}\\&=1+\sum_{r=2}^n \frac{1}{r-1}-\frac{1}{r}\\&=2-\frac{1}{n}\\\implies s_n&<2 \quad \forall n \in \mathbb{N} \end{align}\]

Now, \[1=s_1<s_{2016}<2\\\implies \large \boxed{\lfloor s_{2016} \rfloor=1}\]

Note:\[s_{2016}<\lim_{n \to \infty} s_n = \frac{\pi^2}{6} < 1.645\]

Refer here for the proof of this. – Deeparaj Bhat · 9 months ago

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– Ayush Rai · 9 months ago

Try to answer other problems of the set.You are tooooo good in solving problems.Log in to reply

– Deeparaj Bhat · 9 months ago

Will try after jee advanced :)Log in to reply

– Ayush Rai · 9 months ago

How much did you get in jee-main?Log in to reply

– Deeparaj Bhat · 9 months ago

Not so good... 250.Log in to reply

– Ayush Rai · 9 months ago

That is pretty good. well i have a long way to go.I'm still in 9th std going to the 10th.Log in to reply

– Ayush Rai · 9 months ago

Nice solution..+1.I guessed it anywayLog in to reply

This is again for NMTC lvl 2 2016 – Ankit Kumar Jain · 9 months ago

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– Ayush Rai · 9 months ago

no it is of 2015Log in to reply

– Ankit Kumar Jain · 9 months ago

Oh sorry I meant the previous year IE 2015....I know because I appeared...Log in to reply

– Ayush Rai · 9 months ago

how much problems could you solve??Log in to reply

– Ankit Kumar Jain · 9 months ago

1 and a half....definitely not so good...One of my friends solved about 3 and qualifiedLog in to reply

– Ayush Rai · 9 months ago

even I appearedLog in to reply

– Ankit Kumar Jain · 9 months ago

How many could you solve??Log in to reply

– Ayush Rai · 9 months ago

same as you...1 and a halfLog in to reply

I think the answer is 2. – Ayush Rai · 9 months ago

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