BdMO 2020 Higher Secondary Regional P9

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saifmd
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BdMO 2020 Higher Secondary Regional P9

Unread post by saifmd » Tue Mar 23, 2021 2:20 pm

f(n)=f(n-1)+2n-2 and f(1)=2 then what is the value of f(100) ?

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: BdMO 2020 Higher Secondary Regional P9

Unread post by Mehrab4226 » Tue Mar 23, 2021 7:14 pm

This solution kinda gets messy for generalizing it. But bear with me.
$f(n)=f(n-1)+2n-1$
$f(n)=f(n-1)+2(n-1)$
Let us just focus on the right hand side.
$f(n-1)+2(n-1)$
$=f(n-2)+2(n-2)+2(n-1)$
$=f(n-2)+2((n-2)+(n-1)$
$=f(n-3)+2((n-3)+(n-2)+(n-1))$
This process goes on,
$=f(n-k)+2((n-k)+(n-k+1)+\cdots +(n-2)+(n-1))$
Now to find the value of $f(n)$ , we can only use $f(1)=2$. So to get f(1) on the right hand side we need to get $k=n-1$
$=f(n-n+1)+2((n-n+1)+(n-n+2)+\cdots +(n-2)+(n-1))$
$\therefore f(n)=f(1)+2(1+2+3+ \cdots +n-2+n-1)$[Note that this makes our problem very easy]
So,
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 98}{2})$
$f(100)=2+2(4851)$
$\therefore f(100)=9704$

The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Noor_here
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Joined:Mon Mar 15, 2021 5:10 pm

Re: BdMO 2020 Higher Secondary Regional P9

Unread post by Noor_here » Thu Mar 25, 2021 5:18 pm

Mehrab4226 wrote:
Tue Mar 23, 2021 7:14 pm
This solution kinda gets messy for generalizing it. But bear with me.
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 98}{2})$
$f(100)=2+2(4851)$
$\therefore f(100)=9704$
Just a little correction :)
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 100}{2})$
$f(100)=2+2(4950)$
$\therefore f(100)=9902$

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: BdMO 2020 Higher Secondary Regional P9

Unread post by Mehrab4226 » Fri Mar 26, 2021 2:27 pm

Noor_here wrote:
Thu Mar 25, 2021 5:18 pm
Mehrab4226 wrote:
Tue Mar 23, 2021 7:14 pm
This solution kinda gets messy for generalizing it. But bear with me.
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 98}{2})$
$f(100)=2+2(4851)$
$\therefore f(100)=9704$
Just a little correction :)
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 100}{2})$
$f(100)=2+2(4950)$
$\therefore f(100)=9902$
Oh yes. Yes. Yes. I forgot the formula was $\frac{n(n+1)}{2}$. I thought that it was $\frac{n(n-1)}{2}$ Thanks for pointing that out! :)
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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