Question

If S1,S2, S3 , .... , Sm are the sums of n terms of m A.P.’s whose first terms are 1,2,3,...,mand whose common differences are 1, 3, 5,..., (2m − 1) respectively, then show that S1+S2+S3+......Sm =1/2mn(mn+1)​

Answer 1

Step-by-step explanation:

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]2n ×[2a+(n−1)d]

So,

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

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

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

Similarly;

.

.

.

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

s1 + s2 + s3 + .....sm = \begin{gathered}\frac{n}{2}  \times [ 2 + (n-1) + 4 + 3(n-1)\\ 6 +5(n-1)  + 2m + (n-1)(2m-1)]\end{gathered}2n ×[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)]2n×[(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)]2n×[m(1+m)+m2(n−1)]

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