Given,
[1]+[2]+[3]+.............+[120]
Now we now that,
[x]=1,whenx∈[1,4), [x]=2,whenx∈[4,9) and similarly [x]=10,whenx∈[100,120)
Now using the above formula we get,
E=1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+4+4+…
⇒E=3×1+5×2+7×3+…..+19×9+10×21
⇒E=r=1∑10(2r+1)r
⇒E=2r=1∑10(r2+r)
⇒E=2[610×11×21]+210×11
⇒E=770+55
⇒E=825