Sk=k1+2+3+…+k=2kk(k+1)=2k+1.........(i)
∴S12+S22+....+S102=(21+1)2+(22+1)2+...+(210+1)2 (using equation (i))
=(22)2+(23)2+...+(211)2
=221[22+32+...+112]
=41[(12+22+32+...+112)−12]
=41[611×12×23−1] (using the identity \displaystyle \sum {n}^{2}_{n=1}^{n=n}=\frac{(n)(n+1)(2n+1)}{6})
=41×505
Comparing with the given condition in question, we get
125A=4505
⇒A=5505×3=303.