Let a₁, a₂, a₃, be an arithmetic progression with first term a₁ = 4 and common difference d = 2 . If S_n = _ k=1 ^ n 1 a_k a_ k+1 + a_ k+1 a_k , then the value of _ n S_n is

Step-by-step solution

The general term of the series can be simplified by factoring out a_k a_ k+1 from the denominator: 1 a_k a_ k+1 + a_ k+1 a_k = 1 a_k a_ k+1 ( a_k + a_ k+1 ) Rationalising the expression by multiplying the numerator and denominator by ( a_ k+1 - a_k ) gives: a_ k+1 - a_k a_k a_ k+1 (a_ k+1 - a_k) Since a_ k+1 - a_k = d , the term becomes: a_ k+1 - a_k d a_k a_ k+1 = 1 d ( 1 a_k - 1 a_ k+1 ) Now, summing from k=1 to n , we get a telescoping series: S_n = 1 d ( ( 1 a₁ - 1 a₂ ) + ( 1 a₂ - 1 a₃ ) + + ( 1 a_n - 1 a_ n+1