Find the sum ∑ i = 1 n ∑ j = 1 i ∑ k = 1 j 1 .

Sum of series S n = ∑ i = 1 n ∑ j = 1 i ∑ k = 1 j 1 ⇒ S n = ∑ i = 1 n ∑ j = 1 i ( j ) ⇒ S n = ∑ n n i ( i + 1 ) 2 ⇒ 1 2 ∑ i = 1 n i 2 + i = 1 2 n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ⇒ n ( n + 1 ) 4 ( 2 n + 1 ) 3 + 1 = n ( n + 1 ) 4 ( 2 n + 1 ) + 3 3 = n ( n + 1 ) 4 ( 2 n + 4 ) 3 S n = n ( n + 1 ) ( n + 2 ) 6

Find the sum ∑ i = 1 n ∑ j = 1 i ∑ k = 1 j 1 .

Sum of series S n = ∑ i = 1 n ∑ j = 1 i ∑ k = 1 j 1 ⇒ S n = ∑ i = 1 n ∑ j = 1 i ( j ) ⇒ S n = ∑ n n i ( i + 1 ) 2 ⇒ 1 2 ∑ i = 1 n i 2 + i = 1 2 n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ⇒ n ( n + 1 ) 4 ( 2 n + 1 ) 3 + 1 = n ( n + 1 ) 4 ( 2 n + 1 ) + 3 3 = n ( n + 1 ) 4 ( 2 n + 4 ) 3 S n = n ( n + 1 ) ( n + 2 ) 6

If there be ' m ' A.P's beginning with unity whose common difference is 1 , 2 , 3 … . m . Show that the sum of their n m terms is ( m / 2 ) ( m n - m + n + 1 )

The first A.P in which common difference is 1 ; a n 1 = 1 + ( n - 1 ) ; The second A.P in which common difference is 2 a n 2 = 1 + ( n - 1 ) × 2 The m t h A.P in which common difference is m a n m = 1 + ( n - 1 ) × m So, S = ∑ r = 1 n ( a n 1 + a n 2 + . . . . + a n m ) S = ( 1 + 1 … … + m times ) + ( n - 1 ) ( 1 + 2 + … . m times ) ⇒ S = m + ( n - 1 ) · m ( m + 1 ) 2 ⇒ S = m 2 2 + ( n - 1 ) ( m + 1 ) = m 2 2 + n m - m + n - 1 ⇒ S = m 2 m n - m + n + 1

If a 1 , a 2 , a 3 … a n are in H.P., then prove that a 1 a 2 + a 2 a 3 + … … + a n - 1 a n = ( n - 1 ) a 1 a n

Given, a 1 , a 2 . . . . a n are in H.P So, 1 a 1 , 1 a 2 , 1 a 3 … … · 1 a n are in A.P common difference of A.P; d = 1 a 2 - 1 a 1 = 1 a 3 - 1 a 2 … … So, a 1 a 2 = a 1 - a 2 d , a 2 a 3 = a 2 - a 3 d So, a 1 a 2 + a 2 a 3 + … … + a n - 1 a n = = 1 d a 1 - a 2 + a 2 - a 3 + … … + a n - 1 - a n = 1 d a 1 - a n Since, 1 a n = 1 a 1 + ( n - 1 ) d ⇒ d = a 1 - a n n - 1 . . . 2 By ( 1 ) & ( 2 ) a 1 a 2 + a 2 a 3 + … … … = ( n - 1 ) a 1 a n

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