Find the sum to n terms of each of the series in Exercises 1 to 7.1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Sure, I can help with that. The given series is a product of two consecutive natural numbers. We can write the nth term as n*(n+1).

The sum of the first n terms (S) of the series can be found using the formula for the sum of the first n natural numbers and the sum of the squares of the first n natural numbers.

The sum of the first n natural numbers is given by the formula n*(n+1)/2 and the sum of the squares of the first n natural numbers is given by the formula n*(n+1)*(2n+1)/6.

So, the sum of the first n terms of the given series is:

S = 12 + 23 + 34 + ... + n(n+1) = (1 + 2 + 3 + ... + n) + (1^2 + 2^2 + 3^2 + ... + n^2) = n*(n+1)/2 + n*(n+1)(2n+1)/6 = n(n+1)/2 * (1 + (2n+1)/3) = n*(n+1)*(n+2)/6

So, the sum of the first n terms of the series 12 + 23 + 34 + 45 + ... is n*(n+1)*(n+2)/6.