# Answer to Question #25532 in Algebra for purna chandra sekhar

Question #25532

find nth term of series 4,11,22,37,56

Expert's answer

we see that

a1=4

a2=4+7

a3=4+7+11

a4=4+7+11+15

a5=4+7+11+15+19

so a_n=a_(n-1)+4n-1=a_(n-2)+4*(n-1)-1+4*n-1=....=a1+4*2-1+4*3-1+....+4*n-1=

=a1+4*( 2+3+4+...+n) -(n-1)=a1+4*(2+n)*(n-1)/2-(n-1)=a1+2*(n+2)*(n-1)-(n-1)= =a1+(n-1)(

2(n+2)-1)=a1+(n-1)*(2n+3)=4+(n-1)(2n+3)

Hence a_n=4+2n^2+n-3=2n^2+n+1

a1=4

a2=4+7

a3=4+7+11

a4=4+7+11+15

a5=4+7+11+15+19

so a_n=a_(n-1)+4n-1=a_(n-2)+4*(n-1)-1+4*n-1=....=a1+4*2-1+4*3-1+....+4*n-1=

=a1+4*( 2+3+4+...+n) -(n-1)=a1+4*(2+n)*(n-1)/2-(n-1)=a1+2*(n+2)*(n-1)-(n-1)= =a1+(n-1)(

2(n+2)-1)=a1+(n-1)*(2n+3)=4+(n-1)(2n+3)

Hence a_n=4+2n^2+n-3=2n^2+n+1

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