Question

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Guys!
solve this!
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Answer 1

Step-by-step explanation:

[ (1 x 22) +(2 x 3?)+ (3 x 42) + n( n + 1)1/ + n? x (n +

[12 x 2 + 22 x 3 +3? x 4 + 1)] = (3n + 5)/(3n+ 1)

For numerator,

Tn = n(n + 1)

= n( n? + 2n + 1) = = n° + 2n2 + n

now, An = ETn = E(n3 + 2n? + n)

=

Σn3 + 2Σn? + Ση

we know,

En3 = [n(n + 1)/2]? =

En? = n(n + 1)(2n + 1)/6

En = n(n + 1)/2 use this here,

= [n(n + 1)/2]2 +2n(n + 1)(2n + 1)/

6 + n(n + 1)/2

= n(n + 1)/2 [ n(n + 1)/2 + 2n(2n +1)/3+1] =n(n + 1)/2[ {3n(n + 1) + 4(2n + 1) + 6/6]

=n(n + 1)/2[[3n² + 3n + 8n +4 +6}/6]

= n(n + 1)(3n? + 11n + 10)/12 =

n(n +1)(3n2 + 6n + 5n + 10)/12

=n(n + 1)(3n + 5)(n + 2)/12

again, for denominator

T N = n°(n + 1) = = n° + n?

now, S' n = ET'n

= E(n3 + n) = En? + En?

= = [n(n + 1)/2] + n(n + 1)(2n + 1)/6

= n(n + 1)/2[n(n + 1)/2 + (2n + 1)/3 ] =n(n + 1)/2[[3n(n + 1)+2(2n + 1)/6]

=n(n + 1)/2[[3n2 + 3n + 4n + 2)/6]

= n(n + 1)(3n2 + Zn + 2)/12 =n(n + 1)(3n2 6n + n + 2)/12

= n(n + 1)(3n + 1)(n + 2)/12

=

now, required sum of the series

(1)

_(2)

T/T N = [n(n + 1)(3n + 5)(n + 2)/12]/[n(n + 1)(3n + 1)(n + 2)/12]

(3n + 5)/(3n + 1) =

hence proved

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