 Hello and welcome to the session. In this session we discuss the following question which says find the sum 2 n times and hence find the sum 2 infinity of the series 1 plus 2 into 1 upon 3 plus 3 into 1 upon 3 square plus 4 into 1 upon 3 cube plus and so on. Consider the arithmetical geometric series a plus, a plus d this whole into r plus a plus 2d this whole into r square plus plus 3d whole into r cube plus and so on minus 1 into d and this whole into r to the power m minus 1 plus and so on. The arithmetical geometric series is a series each term of which is found by multiplying the corresponding terms of an AP, a GP. Where we have the AP as a plus d plus a plus 2d plus a plus 3d plus and so on plus a plus m minus 1 into d the GP is 1 plus plus r cube plus r to the power m minus 1 the GP and the corresponding terms of this AP and this GP we get the series that is arithmetical geometric series. So the key idea that we use for this question. Let's now move on to the solution. We are given the series that is the series and we are supposed to find the sum of the series to n terms and also the sum to infinity. The given series 1 plus 2 into 1 upon 3 plus 3 into 1 upon 3 square plus you can see that this is a trig series of the series that is s n is equal to 1 plus 2 into 1 upon 3 plus 3 into 1 upon 3 square 2 1 upon 3 cube plus and so on plus plus 3 plus 4 plus and so on into the GP 1 plus 1 upon 3 plus 1 upon 3 square plus 1 upon 3 cube plus plus and so on would be 1 upon 3 to the power m minus 1 of the GP 1 plus 1 upon 3 plus 1 upon 3 square 3 cube plus 1 upon 3 to the power m minus 1 here the common ratio r is equal to 1 upon 3 I have to apply both sides by 1 upon 3 so 1 upon 3 into s m is equal to into 1 becomes 1 upon 3 and we write 1 upon 3 here plus 1 upon 3 becomes 2 into 1 upon 3 square plus 3 square becomes 3 cube into 1 upon 3 and that becomes 1 upon 3 and that becomes 1 upon 3 to the power m and 1 upon 3 s m we would get 1 upon 3 this whole into s m is equal to 1 upon 3 square 1 upon 3 cube plus 1 upon 3 to the power n minus 1 1 upon 3 to the power n this is further equal to 2 upon 3 s m is equal to 1 plus 1 upon 3 square plus 1 upon 3 cube minus 1 minus 1 upon 3 to the power n you can see that this forms a geometric series the sum of the series is equal to a that is the first term which is 1 into 1 minus which is 1 upon 3 power n and this whole upon 1 upon 3 and so this is equal to into 1 minus 1 upon 3 to the power n so the sum of these is 3 upon 2 into 1 minus 1 upon 3 to the power n the whole so we get 2 upon 3 s m is equal to 1 upon 3 to the power n is equal to 9 by 4 into 1 minus 1 upon 3 to the power n the whole minus 3 upon 2 into 1 upon 3 to the power is equal to 9 upon 4 into 1 minus 1 upon 3 to the power n minus 3 upon 2 into n into 1 upon 3 to the power n the whole is equal to 1 upon 3 to the power n minus 3 upon 2 into n into 1 upon 3 to the power n now here again is equal to 1 upon 3 at 1 upon 3 is equal to series to infinity that is s infinity is equal to limit 9 upon 4 into n the whole into 1 upon 3 to the power it is 1 upon 3 to the power n is 0 and 4 into 1 minus 0 minus 1 upon 3 to the power n is 0 so n into 1 upon 3 to the power n would also be 0 thus this is equal to 9 upon 4 that is s infinity is equal to 9 upon 4 so we have the series infinity is equal to 9 upon 4 thus that is the sum of the series to infinity of the series to n terms to the solution of this question