Prove that the sum of the infinite series 1 · 3 2 + 3 · 5 2 2 + 5 · 7 2 3 + 7 · 9 2 4 + … … … . ∞ = 23 .

We have, S = 1 · 3 2 + 3 · 5 2 2 + 5 · 7 2 3 + 7 · 9 2 4 + … … … . ∞ . . . i S 1 2 = 1 · 3 2 2 + 3 · 5 2 3 + 5 · 7 2 4 + 7 · 9 2 5 + … … … . ∞ . . . ii Subtracting ( i ) & ( ii ) gives S 2 = 1 · 3 2 + 3 · 4 2 2 + 4 · 5 2 3 + . . . … ∞ … iii S 4 = 1 · 3 2 2 + 3 · 4 2 3 + 4 · 5 2 4 + . . . … ∞ … iv Subtracting ( iii ) & ( iv ) gives S 4 = 1 · 3 2 + 3 · 3 2 2 + 4 · 2 2 3 + 4 · 2 2 4 … … … . . ∞ ⇒ S 4 = 3 2 + 9 2 2 + 4 × 2 2 3 1 1 - 1 2 ⇒ S 4 = 3 2 + 9 4 + 2 ⇒ S 4 = 23 4 ⇒ S = 23

Find the sum to n terms: 1 x + 1 + 2 x ( x + 1 ) ( x + 2 ) + 3 x 2 ( x + 1 ) ( x + 2 ) ( x + 3 ) + … … .

Let S = 1 x + 1 + 2 x ( x + 1 ) ( x + 2 ) + 3 x 2 ( x + 1 ) ( x + 2 ) ( x + 3 ) + … … . = 1 - x x + 1 + 2 x ( x + 1 ) ( x + 2 ) + 3 x 2 ( x + 1 ) ( x + 2 ) ( x + 3 ) + … . . . . = 1 - x x + 1 1 - 2 x + 2 + 3 x 2 ( x + 1 ) ( x + 2 ) ( x + 3 ) + … . . . . = 1 - x 2 ( x + 1 ) ( x + 2 ) + 3 x 2 ( x + 1 ) ( x + 2 ) ( x + 3 ) + … . . . . = 1 - x 2 ( x + 1 ) ( x + 2 ) 1 - 3 x + 3 + . . . . . . . . = 1 - x 3 ( x + 1 ) ( x + 2 ) ( x + 3 ) + . . . . . . . . ∴ S = 1 - x n ( x + 1 ) ( x + 2 ) ( x + 3 ) … . . ( x + n )

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

Let S = a 1 1 + a 1 + a 2 1 + a 1 1 + a 2 + a 3 1 + a 1 1 + a 2 1 + a 3 + … … . ⇒ S = 1 1 + a 1 a 1 + a 2 1 + a 2 + a 3 1 + a 1 1 + a 2 1 + a 3 + . . . . . . ⇒ S = a 1 + a 2 + a 1 a 2 1 + a 1 1 + a 2 + a 3 1 + a 1 1 + a 2 1 + a 3 + a 4 1 + a 1 1 + a 2 1 + a 3 1 + a 4 + . . . . . . . ⇒ S = 1 1 + a 1 1 + a 2 a 1 + a 2 + a 1 a 2 + a 3 1 + a 3 + a 4 1 + a 1 1 + a 2 1 + a 3 1 + a 4 + . . . . . . . ⇒ S = a 1 + a 2 + a 3 + a 1 a 2 + a 2 a 3 + a 3 a 1 + a 1 a 2 a 3 1 + a 1 1 + a 2 1 + a 3 + a 4 1 + a 1 1 + a 2 1 + a 3 1 + a 4 + . . . . . . . ⇒ S = a 1 + a 2 + a 3 + . . . + a n + a 1 a 2 + a 2 a 3 + a 3 a 4 + . . . + a n a 1 + a 1 a 2 a 3 . . . a n 1 + a 1 1 + a 2 1 + a 3 . . . . . . 1 + a n ⇒ S = ∑ i = 1 n a i + ∑ i = 1 n - 1 a i a i + 1 + a n a 1 + ∏ i = 1 n a i ∏ i = 1 n 1 + a i

Prove that the number 4 4 4 … … … . . 4 ⏟ n digits 8 8 8 … … … … . 8 ⏟ ( n -1) digits 9 is a perfect square of the number 6 6 6 … … … … . 6 ⏟ 7 ( n -1) digits

We have, LHS = 4 4 4 … … … . . 4 ⏟ n digits 8 8 8 … … … … . 8 ⏟ ( n -1) digits 9 = 9 + 8 × 10 + 8 × 10 2 … . . + 8 × 10 n - 1 + 4 × 10 n + 4 × 10 n + 1 + 4 × 10 n + 2 + … . . + 4 × 10 2 n - 1 = 9 + 8 × 10 10 n - 1 - 1 9 + 4 × 10 n 10 n - 1 9 = 81 + 8 × 10 n - 80 + 4 × 10 2 n - 4 × 10 n 9 = 1 + 4 × 10 n + 4 × 10 2 n 9 = 1 + 2 × 10 n 3 2 RHS = 7 + 6 × 10 10 n - 1 - 1 9 = 63 + 6 × 10 n - 60 9 = 3 2 × 10 n + 1 9 = 1 + 2 × 10 n 3 LHS = RHS Hence, proved

If a , b , c are three positive real number then prove that : a b + c + b a + c + c a + b ≥ 3 2

Given: a , b , c are three positive real numbers. To prove: a b + c + b a + c + c a + b ≥ 3 2 As, A M ≥ H M Using concept of mean of three numbers a b + c + b a + c + c a + b 3 ≥ 3 b + c a + a + c b + a + b c a b + c + b a + c + c a + b 3 ≥ 3 b a + a b + c a + a c + c b + b c . . . . ( 1 ) As, A M ≥ G M using the concept of mean of two numbers Also, b a + a b 2 ≥ a b · b a ∴ b a + a b ≥ 2 , Similarly, c a + a c ≥ 2 , c b + b c ≥ 2 ∴ b a + a b + c a + a c + c b + b c ≥ 6 ∴ 3 b a + a b + c a + a c + c b + b c ≤ 3 6 . . . . ( 2 ) from 1 s t and 2 n d a b + c + b a + c + c a + b ≥ 3 2 .