Find the sum 1 + 2 a + 3 a 2 + … … to n terms.

Let the sum of above series be represented by S and the last term be n a n - 1 . ∴ We can write, S = 1 + 2 a + 3 a 2 + 4 a 3 + … … + n a n - 1 . . . ( i ) This seems to be an arithmetic-geometric progression whose common difference of arithmetic progression (check the coefficients → 1 , 2 , 3 , 4 . . . . . ) is 1 and the common ratio of the geometric progression (check for the variables → 1 , a , a 2 , a 3 , . . . . . . ) is a . To find the sum, let us multiply both sides by the common ratio a as S · a = a + 2 a 2 + 3 a 3 + … … + ( n - 1 ) a n - 1 + n a n . . . ( ii ) Subtracting ( ii ) from ( i ) S ( 1 - a ) = 1 + a + a 2 + a 3 + … + a n - 1 - n a n Applying formula for sum of n terms of a geometric progression in the R.H.S, we get 1 - a S = 1 a n - 1 a - 1 - n a n ∴ S = a n - 1 ( a - 1 ) 2 - n a n 1 - a

Find the sum of the sequence 1 + 3 4 + 7 16 + 15 64 + 31 256 + … … … to infinity.

Let S = 1 + 3 4 + 7 16 + 15 64 + 31 256 + … … ∞ . . . ( i ) Since the sequence is neither in arithmetic or geometric progression, let us try to apply the concept of arithmetic-geometric progression here. It can be noticed that the denominator of the sequence is following a geometric progression with common ratio as 1 4 . Let us multiply both sides by the common ratio 1 4 1 4 S = 1 4 + 3 16 + 7 64 + 15 256 + … … ∞ . . . ( ii ) Subtracting ( ii ) from second terms of ( i ) , we get 1 - 1 4 S = 1 + 2 4 + 4 16 + 8 63 + 16 256 + … … ∞ ⇒ 3 4 S = 1 + 1 2 + 1 4 + 1 8 + 1 16 + … … ∞ Again it can be seen that obtained sequence is an infinite geometric progression with common ratio as 1 2 . Applying formula for sum of infinite geometric progression, we get 3 4 S = 1 1 - 1 2 = 2 ⇒ S = 8 3 .

Find the sum 1 + 3 x + 5 x 2 + 7 x 3 + 9 x 4 + … … . up to infinity, where | x | < 1 .

Let S = 1 + 3 x + 5 x 2 + 7 x 3 + 9 x 4 + … … ∞ . . . ( i ) Multiplying both sides by x and shifting by one term, we get or ⇒ x S = x + 3 x 2 + 5 x 3 + 7 x 4 + ⋯ ⋯ ∞ . . . ii Subtracting ( ii ) from ( i ) by shifting one term in this way ⇒ S - x S = 1 + 2 x + 2 x 2 + 2 x 3 + … … ∞ ⇒ ( 1 - x ) S = 1 + 2 x 1 + x + x 2 + … … ∞ ⇒ ( 1 - x ) S = 1 + 2 x · 1 1 - x ⇒ ( 1 - x ) S = 1 + x 1 - x ⇒ S = 1 + x 1 - x 2 .

Find the sum 1 + 2 2 + 3 2 2 + 4 2 3 + … … up to n terms.

Let S = 1 + 2 2 + 3 2 2 + 4 2 3 + … … . + n 2 n - 1 . . . ( i ) Multiplying both sides by 1 2 and shifting by one term, we get 1 2 S = 0 + 1 2 + 2 2 2 + 3 2 3 + … … n 2 n . . . ( ii ) Subtracting ( ii ) from ( i ) , we get S - 1 2 S = 1 + 1 2 + 1 2 2 + 1 2 3 + … … 1 2 n - 1 - n 2 n 1 2 S = 1 + 1 2 + 1 2 2 + 1 2 3 + . . . . . 1 2 n - 1 - n 2 n The series inside bracket is a geometric progression, having common ratio, r = 1 2 Now, the sum for n terms of a geometric progression having first term a and common ratio r is S n = a 1 - r n 1 - r Therefore, 1 2 S = 1 - 1 2 n 1 - 1 2 - n 2 n ⇒ S = 4 1 - 1 2 n - n 2 n - 1

Find the sum 1 + 3 2 + 5 4 + 7 8 + … … to infinity.

Let S = 1 + 3 2 + 5 4 + 7 8 + … … ∞ . . . i Multiplying both sides by 1 2 and shifting by one term, we get ∴ 1 2 S = 0 + 1 2 + 3 4 + 5 8 + … … ∞ . . . ( ii ) Subtracting ( ii ) from ( i ) , we get 1 2 S = 1 + 2 2 + 2 4 + 2 8 + … … ∞ ⇒ 1 2 S = 1 + 1 + 1 2 + 1 4 + … … ∞ ⇒ 1 2 S = 1 + 1 1 - 1 2 ⇒ 1 2 S = 1 + 2 ⇒ S = 3 × 2 = 6

Find the sum of the series 1 + 3 x + 6 x 2 + 10 x 3 + … … up to infinity, where | x | < 1 .

Given series is an A . G . P . , convert the series into a G . P . by multiplying the series by the common ratio and subtracting from the original series. Let S = 1 + 3 x + 6 x 2 + 10 x 3 + … … ∞ . . . ( i ) Multiplying both sides by x , then we get x · S = x + 3 x 2 + 6 x 3 + 10 x 4 + … … ∞ Or, x · S = 0 + x + 3 x 2 + 6 x 3 + 10 x 4 + … … ∞ . . . ( ii ) Subtracting ( ii ) from ( i ) , we get ( 1 - x ) S = 1 + 2 x + 3 x 2 + 4 x 3 + … … ∞ . . . ( iii ) Multiplying again both sides of ( iii ) by x , we get x ( 1 - x ) S = x + 2 x 2 + 3 x 3 + 4 x 4 + … … ∞ Or, x ( 1 - x ) S = 0 + x + 2 x 2 + 3 x 3 + 4 x 4 + … … ∞ . . . ( iv ) Subtracting ( iv ) from ( iii ) , we get S 1 - x 2 = 1 + x + x 2 + x 3 + . . . . . . ∞ . . . v Since, it is given that, x < 1 and we know that, 1 + x + x 2 + x 3 + . . . . . . . ∞ = 1 1 - x So, from equation ( v ) , we can write S 1 - x 2 = 1 1 - x S = 1 ( 1 - x ) 3 .