 Hello and welcome to the session. In this session we discuss the following question which says sum up to n terms the series 1 cube upon 1 plus 1 cube plus 2 cube upon 1 plus 3 plus 1 cube plus 2 cube plus 3 cube upon 1 plus 3 plus 5 plus and so on. Before we move on to the solution let's discuss very important result to be used in this question which says that if we have the nth term of the series as dn and this is equal to say a n cube plus b n square plus c n plus d then the sum of the n terms of the series given by Sn would be equal to a into summation n cube plus b into summation n square plus c into summation n plus d into n. Then next we have summation n is equal to n into n plus 1 d whole and this whole upon 2 then summation n square is equal to n into n plus 1 d whole into 2 n plus 1 d whole and this whole upon 6 then summation n cube is equal to n into n plus 1 d whole the whole upon 2 and this whole square and this is equal to summation n the whole square since we know that submission n is n into n plus 1 the whole and the whole upon 2. This is the key idea that we use for this question. Next now proceed with the solution. The given series is 1 cube upon 1 plus 1 cube plus 2 cube upon 1 plus 3 plus 1 cube plus 2 cube plus 3 cube this whole upon 1 plus 3 plus 5 plus and so on and we are supposed to find the sum of the n terms of the series for this first of all we need to find the nth term of the series. Now the nth term of the series given by Tn would be equal to the nth term of 1 cube 1 cube plus 2 cube 1 cube plus 2 cube plus 3 cube and so on upon the nth term of 1 1 plus 3 1 plus 3 plus 5 and so on. So we have Tn is equal to 1 cube plus 2 cube plus 3 cube plus n so on plus n cube and this whole upon 1 plus 3 plus 5 plus n so on plus n this means we have Tn is equal to submission n cube upon. Now as you can see that this is an AP so we will find the sum of n terms of this AP that is we have 1 plus 3 plus 5 plus and so on plus n which is the AP here the first term A is 1 and the common difference D is 2. So now sum of the n terms of this AP would be equal to n upon 2 into 2A that is 2 into 1 which is 2 plus n minus 1 into D which is 2 so this is equal to n upon 2 into 2 plus 2n minus 2 this 2 and minus 2 cancels and we are left with n square that is the sum of the AP is n square so we write here n square. Now we know that submission n cube is equal to n into n plus 1 D whole and this whole upon 2 D whole square so this would be equal to n into n plus 1 D whole this whole upon 2 D whole square into 1 upon n square so this is equal to n square into n plus 1 D whole square this whole upon 4 into 1 upon n square then this n square n square cancels and so this is equal to n square plus 2n plus 1 this whole upon 4 this is Tn. Now from the key idea we know that if the nth term of the series is given in this form then we can write the sum of the n terms of the series as A into submission n cube plus B into submission n square plus C into submission n plus Dn so sum of n terms of the series given by Sn is equal to 1 upon 4 into submission n square plus 2 into submission n plus 1 into n that is n the whole. Now we will put the values for submission n square submission n now submission n square is given by n into n plus 1 D whole into 2n plus 1 D whole and whole upon 6 so we have Sn is equal to 1 upon 4 into n into n plus 1 D whole into 2n plus 1 D whole and this whole upon 6 plus 2 into submission n which is given by n into n plus 1 D whole and this whole upon 2 so we write here n into n plus 1 D whole this whole upon 2 plus n the whole now this 2 and 2 cancels and now we can take n common so here we have n upon 4 into n plus 1 D whole into 2n plus 1 D whole this whole upon 6 plus n plus 1 D whole plus 1 now taking the LCM here as 6 we get n plus 1 multiplied by 2n plus 1 gives us 2n square plus 3n plus 1 plus 6 into n plus 1 D whole becomes 6n plus 6 plus 6 into 1 becomes 6 so this is equal to n upon 4 into 2n square plus 3n plus 6n is 9n plus 1 plus 6 plus 6 is 13 and this whole upon 6 further multiplying n by this expression in the numerator we get this is equal to 2n cube plus 9n square plus 13n and this whole upon 4 into 6 which is 24 so this is the Sn which is the sum of the n terms of the series the n terms of the series is equal to 2n cube plus 9n square plus 13n and this whole upon 24 so this is our final answer this completes the session hope you have understood the solution of this question