Show that [1 x 2² + 2 x 3² + .... + n x (n + 1)²] / [1² x 2 + 2² x 3 + .... + n² x (n + 1)] = (3n + 5) / (3n + 1)

Solution:

The n^th term of the given series 1 x 2² + 2 x 3² + .... + n x (n + 1)²

aₙ = n (n + 1)² = n³ + 2n² + n

The n^th term of the given series 1² x 2 + 2² x 3 + .... + n² x (n + 1) in denominator

aₙ = n² (n + 1) = n³ + n²

Therefore,

[1 x 2² + 2 x 3² + .... + n x (n + 1)²] / [1² x 2 + 2² x 3 + .... + n² x (n + 1)] = ∑ⁿₖ ₌ ₁(a)ₖ / ∑ⁿₖ ₌ ₁(a)ₖ

= ∑ⁿₖ ₌ ₁(k³ + 2k² + k) / ∑ⁿₖ ₌ ₁(k³ + k²) ....(1)

Using the summation formulas, we will calculator each of the above series.

∑ⁿₖ ₌ ₁(k³ + 2k² + k) = n² (n + 1)²/4 + 2n (n + 1)(2n + 1)/6 + n (n + 1)/2

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

= n (n + 1)/2 [(3n² + 3n + 8n + 4 + 6)/6]

= n (n + 1)/12 [3n² + 11n + 10]

= n (n + 1)/12 [3n² + 6n + 5n + 10]

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

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

Also,

∑ⁿₖ ₌ ₁(k³ + k²) = [n² (n + 1)²]/4 + [n (n + 1)(2n + 1)]/6

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

= n (n + 1)/2 [3n² + 3n + 4n + 2]

= n (n + 1)/12 [3n² + 7n + 2]

= n (n + 1)/12 [3n² + 6n + n + 2]

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

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

From (1), (2) and (3) , we obtain

[1 x 2² + 2 x 3² + .... + n x (n + 1)²] / [1² x 2 + 2² x 3 + .... + n² x (n + 1)]

= {[n (n + 1)(n + 2)(3n + 5)]/12} / {n (n + 1)(n + 2)(3n + 1)/12}

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

Hence proved

---

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 26

Show that [1 x 2² + 2 x 3² + .... + n x (n + 1)²] / [1² x 2 + 2² x 3 + .... + n² x (n + 1)] = (3n + 5) / (3n + 1)

Summary:

We have proved that [1 x 2² + 2 x 3² + .... + n x (n + 1)²] / [1² x 2 + 2² x 3 + .... + n² x (n + 1)] = (3n + 5) / (3n + 1)