Question

[Chapter: Number Theory]
1) Let $\sf{a = \dfrac{1^2}{1} + \dfrac{2^2}{3} + \dfrac{3^2}{5} + \dots + \dfrac{{(1001)}^{2}}{2001}}$, and $\sf{b = \dfrac{1^2}{3} + \dfrac{2^2}{5} + \dfrac{3^2}{7} + \dots + \dfrac{{(1001)}^{2}}{2003}}$. The closest integer of $\sf{(a - b)}$ is ?
$\sf{\\}$
2) Let s, t be positive integers such that $\sf{{7}^{s}}$ divides 400! and $\sf{{3}^{t}}$ divides $\sf{[(3!)!]!}$, then s + t is ?
$\sf{\\}$
[Ans. (1) => 501 and (B) => 422]
Explanation needed.​

Answer 1

1. Given,

$\sf{\longrightarrow a=\dfrac{1^2}{1}+\dfrac{2^2}{3}+\dfrac{3^2}{5}+\,\dots\,+\dfrac{1001^2}{2001}}$

Or,

$\displaystyle\sf{\longrightarrow a=\sum_{r=1}^{1001}\dfrac{r^2}{2r-1}}$

And,

$\sf{\longrightarrow b=\dfrac{1^2}{3}+\dfrac{2^2}{5}+\dfrac{3^2}{7}+\,\dots\,+\dfrac{1001^2}{2003}}$

Or,

$\displaystyle\sf{\longrightarrow b=\sum_{r=1}^{1001}\dfrac{r^2}{2r+1}}$

Then,

$\displaystyle\sf{\longrightarrow a-b=\sum_{r=1}^{1001}\dfrac{r^2}{2r-1}-\sum_{r=1}^{1001}\dfrac{r^2}{2r+1}}$

$\displaystyle\sf{\longrightarrow a-b=\sum_{r=1}^{1001}r^2\left[\dfrac{1}{2r-1}-\dfrac{1}{2r+1}\right]}$

$\displaystyle\sf{\longrightarrow a-b=\sum_{r=1}^{1001}r^2\left[\dfrac{(2r+1)-(2r-1)}{(2r-1)(2r+1)}\right]}$

$\displaystyle\sf{\longrightarrow a-b=\sum_{r=1}^{1001}\dfrac{2r^2}{4r^2-1}\right]}$

$\displaystyle\sf{\longrightarrow a-b=2\sum_{r=1}^{1001}\dfrac{r^2}{4r^2-1}\right]}$

As $\sf{r}$ varies from $\sf{1}$ to $\sf{1001,}$ $\sf{4r^2>>1,}$ then I can neglect that 1 in the denominator so that the value of $\sf{a-b}$ obtained will be slightly less than actual value of $\sf{a-b.}$

So,

$\displaystyle\sf{\longrightarrow a-b\simeq2\sum_{r=1}^{1001}\dfrac{r^2}{4r^2}\right]}$

$\displaystyle\sf{\longrightarrow a-b\simeq2\sum_{r=1}^{1001}\dfrac{1}{4}\right]}$

$\displaystyle\sf{\longrightarrow a-b\simeq\dfrac{2\times1001}{4}\right]}$

$\displaystyle\sf{\longrightarrow a-b\simeq500.5}$

This value is slightly less than actual value.

So the nearest integer would be,

$\displaystyle\sf{\longrightarrow\underline{\underline{a-b\simeq501}}}$

2. $\sf{s}$ is the greatest possible positive integer such that $\sf{7^s\mid 400!.}$

Let [.] be greatest integer function.

Then, since $\sf{7^3<400<7^4,}$

$\sf{\longrightarrow s=\left[\dfrac{400}{7^1}\right]+\left[\dfrac{400}{7^2}\right]+\left[\dfrac{400}{7^3}\right]}$

$\sf{\longrightarrow s=\left[\dfrac{400}{7}\right]+\left[\dfrac{400}{49}\right]+\left[\dfrac{400}{343}\right]}$

$\sf{\longrightarrow s=57+8+1}$

$\sf{\longrightarrow s=66}$

$\sf{t}$ is the greatest possible positive integer such that $\sf{3^t\mid [(3!)!]!=720!.}$

Then, since $\sf{3^5<720<3^6,}$

$\sf{\longrightarrow t=\left[\dfrac{720}{3^1}\right]+\left[\dfrac{720}{3^2}\right]+\left[\dfrac{720}{3^3}\right]+\left[\dfrac{720}{3^4}\right]+\left[\dfrac{720}{3^5}\right]}$

$\sf{\longrightarrow t=\left[\dfrac{720}{3}\right]+\left[\dfrac{720}{9}\right]+\left[\dfrac{720}{27}\right]+\left[\dfrac{720}{81}\right]+\left[\dfrac{720}{243}\right]}$

$\sf{\longrightarrow t=240+80+26+8+2}$

$\sf{\longrightarrow t=356}$

Therefore,

$\sf{\longrightarrow s+t=66+356}$

$\sf{\longrightarrow\underline{\underline{s+t=422}}}$