Write the first five terms of each of the following sequences whose nth terms are:
$ a_{n}=2 n^{2}-3 n+1 $

Given:

\( a_{n}=2 n^{2}-3 n+1 \)

To do:

We have to find the first five terms of the given sequence.

Solution:

$a_n=2n^{2}-3n+1$

Taking $n=1$, we get

$a_1=2(1^{2})-3(1)+1=2(1)-3+1=2-2=0$

Taking $n=2$, we get

$a_2=2(2^{2})-3(2)+1=2(4)-6+1=8-5=3$

Taking $n=3$, we get

$a_3=2(3^{2})-3(3)+1=2(9)-9+1=18-8=10$

Taking $n=4$, we get

$a_4=2(4^{2})-3(4)+1=2(16)-12+1=32-11=21$

Taking $n=5$, we get

$a_5=2(5^{2})-3(5)+1=2(25)-15+1=50-14=36$

Hence, the first five terms of the given sequence are $0, 3, 10, 21$ and $36$.   

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Updated on: 10-Oct-2022

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