If 1+(√3-√2)/(2 √3)+(5-2 √6)/18+(9 √3-11 √2)/(36 √3)+(49-20 √6)/180+… upto ∞=2+(√(b/a)+1) log ₑ(a/b), where a and b are integers with gcd(a, b)=1, then 11 a+18 b is equal to __________.

Solution

Maths sir · 5 steps

Spot that every term is a power of x = 1-√(2/3) over n(n+1), then split it into two log series.

1. Find the base. Put x = (√3-√2)/√3 = 1-√(2/3). Then x² = (5-2√6)/3, x³ = (9√3-11√2)/3√3 and x⁴ = (49-20√6)/9.
2. General term. The terms after 1 are x/2, x²/6, x³/12, x⁴/20, and 2, 6, 12, 20 are n(n+1). Write 1/n(n+1) = 1/n-1/(n+1) and use Σ xⁿ/n = -ln(1-x).

   S = 1+Σₙ₌₁^∞xⁿ/n(n+1) = 1 - ln(1-x) + (ln(1-x)+x)/x = 2+(1/x-1)ln(1-x)

3. Put in x. Here 1-x = √(2/3), so ln(1-x) = 1/2 ln 2/3. Also 1/x-1 = (1-x)/x = √2/(√3-√2) = √6+2.
4. Match the form. S = 2+(√6+2)/2 ln 2/3 = 2+(√(3/2)+1)ln 2/3. Comparing with the given form, b/a = 3/2 and a/b = 2/3, so a = 2 and b = 3.
5. So. 11a+18b = 22+54 = 76.

The common slip: 69. Swapping a and b (taking a = 3, b = 2) gives 69. Check that √(b/a) and ln(a/b) both fit the same pair.

The idea to keep. When the denominators run 2, 6, 12, 20, think n(n+1). Split it into partial fractions and both pieces become the -ln(1-x) series.

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