Question

if s1, s2, s3........sm are the sum of n terms of m AP's whose first terms are 1, 2.3..........m and common differences are 1, 3,5..........(2m-1), then show thats1+s2+s3........sm= mn/2 (mn+1)

Answer 1

See the answer below:-

Answer 2

Answer:

Since , given are the sum of n terms of an AP s1, s2, s3........sm.

The first term are 1 , 2, 3.......m

and common difference are 1 ,3 , 5 ......(2m-1)

To Prove :  s1 + s2 + s3......+sm = mn/2(mn +1).

Proof:

Let us first find the sum of all n terms of the AP;

∵ s = $\frac{n}{2}  \times [2a + (n-1)d]$

So,

s1 = $\frac{n}{2} \times [2 \times1 + (n-1 )1]$

s1 = $\frac{n}{2} \times [ 2 + (n-1)1]$

s2 =  $\frac{n}{2} \times [ 4 + (n-1)3]$

Similarly;

.

.

.

sm =  $\frac{n}{2} \times [ 2m + (n-1)(2m-1)]$

Now , we need to sum the following;

s1 + s2 + s3 + .....sm;

$\frac{n}{2} \times [2 \times1 + (n-1 )1]$  + s2 =  $\frac{n}{2} \times [ 4 + (n-1)3]$ + ....... +  $\frac{n}{2} \times [ 2m + (n-1)(2m-1)]$

s1 + s2 + s3 + .....sm = $\frac{n}{2}  \times [ 2 + (n-1) + 4 + 3(n-1)\\ 6 +5(n-1)  + 2m + (n-1)(2m-1)]$

$\frac{n}{2} \times [(2 + 4 + 6 +......2m) + (n-1) + 3(n-1) +5(n-1) +....+ (n-1)(2m-1)]$

∴  s1 + s2 + s3 + .....sm =  $\frac{n}{2} \times [m (1+m)+ m^{2} (n-1)]$

∴ s1 + s2 + s3 + .....sm  = mn/2(mn +1)          proved.