Question

If S1, S2, S3, ... , SP be the sum of n terms of an AP's whose first terms are 1, 2, 3, ... ,p and common difference s 1, 3, 5, ... , (2p-1) respectively. Show that S1+S2+S3+...+SP=[ np(np+1)]/2.

Answer 1

we know, Sn = n/2 [2a + (n - 1)d ]

so, S1 = n/2 [2 × 1 + (n - 1) × 1]

= n/2[ 2 + n - 1] = n(n + 1)/2

similarly, S2 = n/2 [2 × 2 + (n - 1) × 3]

= n/2 [4 + 3n - 3] = n(3n + 1)/2

S3 = n/2 [2 × 3 + (n - 1) × 5 ]

= n(5n + 1)/2

...........

.........

Sp = [n{(2p - 1)n + 1}]/2

now, LHS = S1 + S2 + S3 + S4 + ... + Sp

= n(n + 1)/2 + n(3n + 1)/2 + n(5n + 1)/2 + .... + [n{(2p - 1)n + 1}]/2

= 1/2 [ n²{1 + 3 + 5 + 7 + .. + (2p - 1) } + n{1 + 1 + 1 + ... p terms }]

we know, 1 + 3 + 5 + 7 + ... + (2p - 1) = p/2[2 × 1 + (p - 1) × 2 ] = p²

= 1/2 [ n²(p²) + np ]

= np(np + 1)/2 = RHS [hence proved ]