The value of lim _(n → ∞) (1 + 2 - 3 + 4 + 5 - 6 + ..... + (3n - 2) + (3n - 1) - 3n)/(√(2n⁴ + 4n + 3) - √(n⁴ + 5n + 4)) is :

Solution

Maths sir · 4 steps

Group the numerator in threes to get its sum, then divide top and bottom by n².

1. Group in threes. The terms come in blocks (3k-2) + (3k-1) - 3k for k = 1, 2, …, n. Each block adds up to 3k - 3.

   Σₖ₌₁ⁿ (3k-3) = 3n(n+1)/2 - 3n = (3n² - 3n)/2

2. Size of denominator. The leading terms √2n⁴ = √2 n² and √n⁴ = n² do not cancel, so you do not need to rationalise. Dividing by n² gives √(2 + 4/n³ + 3/n⁴) - √(1 + 5/n³ + 4/n⁴) → √2 - 1.
3. Divide by n squared. The numerator over n² is 3/2 - 3/2n → 3/2. So the limit is 3/2(√2 - 1).
4. So. Multiply top and bottom by √2 + 1, using (√2 - 1)(√2 + 1) = 1. The limit is 3/2(√2 + 1), option (b).

The idea to keep. For a long patterned sum, find the repeating block and add the blocks. Then compare the highest powers of n on top and bottom. Only the leading coefficients survive.

Answer from the NTA final answer key. Solution by PrepWiser.