C-13-Q123mediumsingle_mcqThe sum of the first 101010 terms of log2+log4+log8+⋯\log 2 + \log 4 + \log 8 + \cdotslog2+log4+log8+⋯ is —a55log255 \log 255log2b45log245 \log 245log2c50log250 \log 250log2d10log210 \log 210log2ব্যাখ্যাWriting each term as log2, 2log2, 3log2,…\log 2,\,2\log 2,\,3\log 2,\dotslog2,2log2,3log2,…, the sum of 101010 terms is log2 (1+2+⋯+10)=log2⋅10⋅112=55log2\log 2\,(1+2+\cdots+10)=\log 2\cdot\dfrac{10\cdot 11}{2}=55\log 2log2(1+2+⋯+10)=log2⋅210⋅11=55log2.