 Hi and welcome to the session. I am Asha and let us solve the following question that says find the sum to n terms of each of the series. Fourth is 1 upon 1 into 2 plus 1 upon 2 into 3 plus 1 upon 3 into 4 plus 4. So let's start with the solution and let the given series be denoted by s. s is 1 upon 1 into 2 plus 1 upon 2 into 3 plus 1 upon 3 into 4 plus so on in the last term. In the nth term will be 1 upon n into n plus 1 of the given series v a k and a k will be equal to 1 upon k into k plus 1. Now we are using partial fraction at a k which is 1 upon k into k plus 1 a upon k plus b upon k plus 1. So this implies 1 is equal to a into k plus 1 plus b into k which further implies 1 is equal to a plus b into k plus a. Now I am comparing we find here that a is equal to 1 and a plus b is equal to 0 which further implies that b is equal to minus 1. Thus we have a is equal to 1 and b is equal to minus 1. So a k can be further written as 1 upon k plus minus 1 upon k plus 1 or 1 upon k plus minus 1 upon k plus 1. So opening the bracket plus into minus is minus this is further equal to 1 upon k minus 1 upon k plus 1. So this is the kth term of the series. So the series is can further be rewritten as first term is a1. So it can be written as 1 upon 1 minus 1 upon 2 this is the first term plus the second term can be written as 1 upon 2 minus 1 upon 3 plus the third term can be written as 1 upon 3 minus 1 upon 4 plus so on as term which is the nth term can be written as 1 upon n minus 1 upon n plus 1. Now let us solve it. So this will be equal to 1 upon 1 minus 1 upon 2 plus 1 upon 2 minus 1 upon 3 plus 1 upon 3 minus 1 upon 4 plus so on up to 1 upon n minus 1 upon n plus 1. So in cancelling we are left with 1 upon 1 minus 1 upon n plus 1 on taking LCM n plus 1 in the denominator n plus 1 minus 1 in the numerator which gives n upon n plus 1. Now to n terms of the given series is n upon n plus 1. So this completes the session. Take care and have a good day.