Given, a_n=( 1)^n 1. 5^n+1 Substituting n= 1, 2, 3, 4, 5 , we obtain #160; a _ 1 = ( 1) ^ 1 1 5 ^ 1+1 = 5 ^ 2 =25 a _ 2 = ( 1) ^ 2 1 5 ^ 2+ ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Monomial

a n = -1 n-1 ...

Question

$a_{n} = {(- 1)}^{n - 1} 5^{n + 1}$

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Solution

Given, $a_{n} = (- 1)^{n - 1} . 5^{n + 1}$
Substituting $n = 1, 2, 3, 4, 5$ , we obtain
$a_{1} = {(- 1)}^{1 - 1} 5^{1 + 1} = 5^{2} = 25 a_{2} = {(- 1)}^{2 - 1} 5^{2 + 1} = {- 5}^{3} = - 125 a_{3} = {(- 1)}^{3 - 1} 5^{3 + 1} = 5^{4} = 625 a_{4} = {(- 1)}^{4 - 1} 5^{4 + 1} = 5^{5} = - 3125 a_{5} = {(- 1)}^{5 - 1} 5^{5 + 1} = 5^{6} = 15625$

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an=(−1)n−15n+1

Given, an=(−1)n−1.5n+1Substituting n=1,2,3,4,5, we obtaina1=(−1)1−151+1=52=25a2=(−1)2−152+1=−53=−125a3=(−1)3−153+1=54=625a4=(−1)4−154+1=55=−3125a5=(−1)5−155+1=56=15625

$a_{n} = {(- 1)}^{n - 1} 5^{n + 1}$