Just make it simple

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Let there be a sequence denoted by $a_{k}$ for $k \ge 0, k \in \mathbb{Z}$ . If:

$a_0=4 \\ a_1=5 \\ a_2=8 \\ a_3=13$

Compute $|a_{20}-a_{18}|$

The answer is 76.

2 solutions

Rishabh Jain
Mar 15, 2016

General term: $a_n=a_{n-1}+2n-1~~,n\geq1$ While we can easily notice : $a_{n+1}-a_{n-1}=4(n)$ Hence, $|a_{20}-a_{18}| =4(19)=\boxed{76}$

Otto Bretscher
Mar 15, 2016

It looks like $a_n-a_{n-1}=2n-1$ so $a_{20}-a_{18}=a_{20}-a_{19}+a_{19}-a_{18}=39+37=\boxed{76}$

Bingo! Same way ;-)..

Rishabh Jain - 5 years, 3 months ago

Simple indeed... but I never miss writing a one-line solution if I can ;)

Otto Bretscher - 5 years, 3 months ago

I knew you would mention that .. ;-)

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – @Rishabh Cool : Now it's your turn to write a one-line solution for this one

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – Tried that already ...... Its beyond my scope and need to learn a lot before writing a 1-liner (or a Method 1,2,3) for that question.. :-)

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – Don't sell yourself short! You can do that one using methods that are well within your reach.

Otto Bretscher - 5 years, 3 months ago

Just make it simple

The answer is 76.

2 solutions

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