Find the sum of the series: 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … to n terms

Let T n be the n th term of given series. Thus, we have T n = n ( n + 1 ) = n 2 + n Now, let S n be the sum of n terms of given series. ∴ S n = ∑ k = 1 n T k ⇒ S n = ∑ k = 1 n k 2 + k ⇒ S n = ∑ k = 1 n k 2 + ∑ k = 1 n k ⇒ S n = n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ⇒ S n = n ( n + 1 ) 2 2 n + 1 3 + 1 ⇒ S n = n ( n + 1 ) 2 2 n + 4 3 ⇒ S n = n ( n + 1 ) ( 2 n + 4 ) 6 ⇒ S n = n ( n + 1 ) ( n + 2 ) 3

Find the sum of the series: 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … to n terms

Let T n be the n th term of given series. Thus, we have T n = n ( n + 1 ) = n 2 + n Now, let S n be the sum of n terms of given series. ∴ S n = ∑ k = 1 n T k ⇒ S n = ∑ k = 1 n k 2 + k ⇒ S n = ∑ k = 1 n k 2 + ∑ k = 1 n k ⇒ S n = n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ⇒ S n = n ( n + 1 ) 2 2 n + 1 3 + 1 ⇒ S n = n ( n + 1 ) 2 2 n + 4 3 ⇒ S n = n ( n + 1 ) ( 2 n + 4 ) 6 ⇒ S n = n ( n + 1 ) ( n + 2 ) 3

Find the sum of the series: 3 × 8 + 6 × 11 + 9 × 14 + … to n terms

The given series is 3 × 8 + 6 × 11 + 9 × 14 + … to n terms = ( nth term of 3 , 6 , 9 … ) × n t h term of 8 , 11 , 14 … = ( 3 n ) ( 3 n + 5 ) = 9 n 2 + 15 n ∴ S n = ∑ k = 1 n a k = ∑ k = 1 n 9 k 2 + 15 k = 9 × n ( n + 1 ) ( 2 n + 1 ) 6 + 15 × n ( n + 1 ) 2 = 3 n ( n + 1 ) 2 ( 2 n + 1 + 5 ) = 3 n ( n + 1 ) 2 ( 2 n + 6 ) = 3 n ( n + 1 ) ( n + 3 )

Find the sum of the series: 1 × 2 2 + 2 × 3 2 + 3 × 4 2 + … to n terms

The given series is 1 × 2 2 + 2 × 3 2 + 3 × 4 2 + … to n terms Now, a n = n ( n + 1 ) 2 ⇒ a n = n 3 + 2 n 2 + n To find: S n = ∑ k = 1 n a k = ∑ k = 1 n k 3 + 2 k 2 + k = ∑ k = 1 n k 3 + 2 ∑ k = 1 n k 2 + ∑ k = 1 n k = n ( n + 1 ) 2 2 + 2 n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 = n ( n + 1 ) 2 n ( n + 1 ) 2 + 2 ( 2 n + 1 ) 3 + 1 = n ( n + 1 ) 2 3 n 2 + 3 n + 8 n + 4 + 6 6 = n ( n + 1 ) 2 3 n 2 + 11 n + 10 6 = n ( n + 1 ) ( 3 n 2 + 5 n + 6 n + 10 ) 12 = n ( n + 1 ) ( n + 2 ) ( 3 n + 5 ) 12

Find the sum of the series: 1 × 2 2 + 3 × 3 2 + 5 × 4 2 + … to n terms

The given series is 1 × 2 2 + 3 × 3 2 + 5 × 4 2 + … to n terms = 1 × ( 1 + 1 ) 2 , 3 × ( 2 + 1 ) 2 … 2 n − 1 × ( n + 1 ) 2 So, a n = ( 2 n − 1 ) ( n + 1 ) 2 = ( 2 n − 1 ) n 2 + 2 n + 1 = 2 n 3 + 4 n 2 + 2 n - n 2 - 2 n - 1 = 2 n 3 + 3 n 2 − 1 ∴ S n = ∑ k = 1 n a k S n = ∑ k = 1 n 2 k 3 + 3 k 2 − 1 = 2 ∑ k = 1 n k 3 + 3 ∑ k = 1 n k 2 − ∑ k = 1 n ( 1 ) = 2 n ( n + 1 ) 2 2 + 3 n ( n + 1 ) ( 2 n + 1 ) 6 − n = n 2 ( n + 1 ) 2 2 + n ( n + 1 ) ( 2 n + 1 ) 2 − n = n 2 n 2 + 1 + 2 n 2 + n 2 n 2 + n + 2 n + 1 2 − n = n 4 + n 2 + 2 n 3 + 2 n 3 + 3 n 2 + n − 2 n 2 = n 4 + 4 n 3 + 4 n 2 − n 2 = n 2 n 3 + 4 n 2 + 4 n − 1 .

Find the sum of the series: 3 × 1 2 + 5 × 2 2 + 7 × 3 2 + … to n terms

Let T n be the n th term of given series. Thus, we have: T n = ( 2 n + 1 ) n 2 = 2 n 3 + n 2 Let S n be the sum of n terms of given series. ∴ S n = ∑ k = 1 n T k ⇒ S n = ∑ k = 1 n 2 k 3 + k 2 ⇒ S n = 2 ∑ k = 1 n k 3 + ∑ k = 1 n k 2 ⇒ S n = 2 n 2 ( n + 1 ) 2 4 + n ( n + 1 ) ( 2 n + 1 ) 6 ⇒ S n = n 2 ( n + 1 ) 2 2 + n ( n + 1 ) ( 2 n + 1 ) 6 ⇒ S n = n ( n + 1 ) 2 n ( n + 1 ) + 2 n + 1 3 ⇒ S n = n ( n + 1 ) 2 3 n 2 + 3 n + 2 n + 1 3 ⇒ S n = n ( n + 1 ) 2 3 n 2 + 5 n + 1 3 ⇒ S n = n ( n + 1 ) 6 3 n 2 + 5 n + 1

Find the sum of the series: ( 1 × 2 × 3 ) + ( 2 × 3 × 4 ) + ( 3 × 4 × 5 ) + … to n terms

The given series is ( 1 × 2 × 3 ) + ( 2 × 3 × 4 ) + ( 3 × 4 × 5 ) + … to n terms Here, a n = n ( n + 1 ) ( n + 2 ) = n 2 + n ( n + 2 ) = n 2 ( n + 2 ) + n ( n + 2 ) = n 3 + 2 n 2 + n 2 + 2 n = n 3 + 3 n 2 + 2 n Let S n be the sum of n terms of given series. ∴ S n = ∑ k = 1 n a k = ∑ k = 1 n k 3 + 3 k 2 + 2 k = ∑ k = 1 n k 3 + 3 ∑ k = 1 n k 2 + 2 ∑ k = 1 n k = n ( n + 1 ) 2 2 + 3 × n ( n + 1 ) ( 2 n + 1 ) 6 + 2 n ( n + 1 ) 2 = n 2 ( n + 1 ) 2 4 + n ( n + 1 ) ( 2 n + 1 ) 2 + n ( n + 1 ) = n ( n + 1 ) 2 n ( n + 1 ) 2 + ( 2 n + 1 ) + 2 = n ( n + 1 ) 2 n 2 + n 2 + 2 n + 1 + 2 = n ( n + 1 ) 2 n 2 + n 2 + 2 n + 3 = n ( n + 1 ) 2 n 2 + n + 4 n + 6 2 = n ( n + 1 ) 2 n 2 + 5 n + 6 2 = n ( n + 1 ) 4 n 2 + 2 n + 3 n + 6 = n ( n + 1 ) 4 [ n ( n + 3 ) + 2 ( n + 3 ) ] = n ( n + 1 ) ( n + 2 ) ( n + 3 ) 4 .

Find the sum of the series: ( 1 × 2 × 4 ) + ( 2 × 3 × 7 ) + ( 3 × 4 × 10 ) + … to n terms

Let T n be the n th term of given series.Thus we have: T n = n ( n + 1 ) ( 3 n + 1 ) = n 3 n 2 + 4 n + 1 = 3 n 3 + 4 n 2 + n Now, let S n be the n term of given series. ∴ S n = ∑ k = 1 n T k ⇒ S n = ∑ k = 1 n 3 k 3 + ∑ k = 1 n 4 k 2 + ∑ k = 1 n k ⇒ S n = 3 ∑ k = 1 n k 3 + 4 ∑ k = 1 n k 2 + ∑ k = 1 n k ⇒ S n = 3 n 2 ( n + 1 ) 2 4 + 4 n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ⇒ S n = 3 n 2 ( n + 1 ) 2 4 + 2 n ( n + 1 ) ( 2 n + 1 ) 3 + n ( n + 1 ) 2 ⇒ S n = n ( n + 1 ) 2 3 n ( n + 1 ) 2 + 4 ( 2 n + 1 ) 3 + 1 ⇒ S n = n ( n + 1 ) 2 3 n 2 + 3 n 2 + 8 n + 4 3 + 1 ⇒ S n = n ( n + 1 ) 2 9 n 2 + 9 n + 16 n + 8 + 6 6 ⇒ S n = n ( n + 1 ) 12 9 n 2 + 25 n + 14

Find the sum of the series: 1 ( 1 × 2 ) + 1 ( 2 × 3 ) + 1 ( 3 × 4 ) + … to n terms

The given series is 1 ( 1 × 2 ) + 1 ( 2 × 3 ) + 1 ( 3 × 4 ) + … to n terms = 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + … + 1 n ⁡ ( n + 1 ) So, a n = 1 n ( n + 1 ) Consider, a n = 1 n ( n + 1 ) = n + 1 - n n n + 1 = 1 n - 1 n + 1 So, a 1 = 1 − 1 2 → ( 1 ) and a 2 = 1 2 − 1 3 → ( 2 ) and a 3 = 1 3 − 1 4 and a n = 1 n − 1 n + 1 Adding above terms column wise, we get a 1 + a 2 + … + a n = 1 1 + 1 2 + 1 3 + … 1 n − 1 2 + 1 3 + 1 4 + … 1 n + 1 ∴ S n = 1 − 1 n + 1 = n + 1 − 1 n + 1 = n n + 1

Find the sum of the infinite series: 8 + 4 2 + 4 + 2 2 + … ∞

We have: 8 + 4 2 + 4 + 2 2 + … ∞ The given G.P has first term, a = 8 and the common ratio r = a 2 a 1 = 4 2 8 = 1 2 Also, r < 1 . Hence, the sum S to ∞ is given by: S ∞ = a 1 − r = 8 1 − 1 2 = 8 2 2 − 1 = 8 2 2 − 1 × 2 + 1 2 + 1 = 8 ( 2 + 2 ) .