BdMO National Secondary 2020 P5

[Mursalin]

এমন কতগুলো বাস্তব সংখ্যা \(x_1, x_2, \cdots\) আছে যেখানে \(n>0\) এর জন্য, \(x_{n+3} = x_{n+2} - 2x_{n+1} + x_n\) হয় । ধরো, \(x_1 = x_3 = 1\) এবং বলা হয়েছে \(x_{98} = x_{99}\) । উপরের শর্ত অনুযায়ী, \(x_1 + x_2 + ... + x_{100} =\) ?


Let \(x_1, x_2, \cdots\) be real numbers so that for all \(n > 0\),
\(x_{n+3} = x_{n+2} - 2x_{n+1} + x_n\). Suppose \(x_1 = x_3 = 1\) and you're given that \(x_{98} = x_{99}\). Find the sum \(x_1 + x_2 + ....+ x_{100}\).

[Anindya Biswas]

We will use telescopic sum:
$x_4=x_3-2x_2+x_1$
$x_5=x_4-2x_3+x_2$
$x_6=x_5-2x_4+x_3$
$\vdots$
$x_{99}=x_{98}-2x_{97}+x_{96}$
$x_{100}=x_{99}-2x_{98}+x_{97}$
Summing, we get $x_4+\cdots+x_{100}=x_1-x_2+x_{99}-x_{98}$
$\therefore x_1+x_2+\cdots+x_{100}=x_1+x_3+x_1+x_{99}-x_{98}=3$.

Another note : The coefficient $1, -2, 1$ gives us the opportunity to apply telescopic sum. Other rows of Pascal's triangle also works well for telescopic sum, for example $1,-3,3,-1$. Binomial coefficients suits nicely with the idea of telescopic summation.