A) \[\frac{1}{2}mn(mn+1)\]
B) \[mn(m+1)\]
C) \[\frac{1}{4}mn(mn-1)\]
D) None of the above
Correct Answer: A
Solution :
Here \[a=1,\ 2,\ 3,\,........,m;\ \ \ d=1,\ 3,\ 5,........,2m-1\] and \[n=n\], then \[{{S}_{1}}+{{S}_{2}}+.......+{{S}_{m}}=\frac{1}{2}mn(mn+1)\] \[\left[ \text{Using}\ S\ =\frac{m}{2}(a+l).\ \text{Since}\ {{S}_{1}},\ {{S}_{2}},\ {{S}_{3}},......{{S}_{m}}\ \text{form}\ \text{an}\ \text{A}\text{.P}\text{.} \right]\]
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