Question

Question 26 Show that [(1 x 2^2) + (2 x 3^2) + ... + (n x (n+1)^2)] / [(1^2 x 2) + (2^2 x 3) + ... + (n^2 x (n+1))] = (3n+5) / (3n+1)

Class X1 - Maths -Sequences and Series Page 200

Answer 1

[ (1 × 2²) +(2 × 3²) + (3 × 4²) + ....... n( n + 1)²]/[1² × 2 + 2² × 3 + 3² × 4 + ......... + n² × (n + 1) ] = (3n + 5)/(3n+ 1)


For numerator ,
Tₙ = n(n + 1)²
= n( n² + 2n + 1)
= n³ + 2n² + n
now, Sₙ = ∑Tₙ
= ∑(n³ + 2n² + n)
= ∑n³ + 2∑n² + ∑n

we know,
∑n³ = [n(n + 1)/2]²
∑n² = n(n + 1)(2n + 1)/6
∑n = 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)(3n² + 11n + 10)/12
= n(n + 1)(3n² + 6n + 5n + 10)/12
=n(n + 1)(3n + 5)(n + 2)/12 ______(1)

again, for denominator
T'ₙ = n²(n + 1)
= n³ + n²
now, S'ₙ = ∑T'ₙ
= ∑(n³ + n²)
= ∑n³ + ∑n²
= [n(n + 1)/2]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 + 7n + 2)/12
=n(n + 1)(3n² + 6n + n + 2)/12
= n(n + 1)(3n + 1)(n + 2)/12 ________(2)

now, required sum of the series
Tₙ/T'ₙ = [n(n + 1)(3n + 5)(n + 2)/12]/[n(n + 1)(3n + 1)(n + 2)/12]
= (3n + 5)/(3n + 1)
hence proved