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algebra intermediate
Problem
Let be the base-10 logarithm of the sum of the elements of the th row in Pascal's triangle. Express in terms of . Recall that Pascal's triangle begins
\begin{array}{rccccccccc} $n=0$:& & & & & 1\\\noalign{\smallskip\smallskip} $n=1$:& & & & 1 & & 1\\\noalign{\smallskip\smallskip} $n=2$:& & & 1 & & 2 & & 1\\\noalign{\smallskip\smallskip} $n=3$:& & 1 & & 3 & & 3 & & 1\\\noalign{\smallskip\smallskip} $n=4$:& 1 & & 4 & & 6 & & 4 & & 1\\\noalign{\smallskip\smallskip} & & & & & $\vdots$ & & & & \end{array}
\begin{array}{rccccccccc} $n=0$:& & & & & 1\\\noalign{\smallskip\smallskip} $n=1$:& & & & 1 & & 1\\\noalign{\smallskip\smallskip} $n=2$:& & & 1 & & 2 & & 1\\\noalign{\smallskip\smallskip} $n=3$:& & 1 & & 3 & & 3 & & 1\\\noalign{\smallskip\smallskip} $n=4$:& 1 & & 4 & & 6 & & 4 & & 1\\\noalign{\smallskip\smallskip} & & & & & $\vdots$ & & & & \end{array}
Solution
Computing the sums of the entries in the first few rows suggestions that the sum of the entries in row is . Indeed, one way to prove this formula is to note that the th entry of the th row is (if we say that the entries in the th row are numbered ). We have since both sides calculate the number of ways to choose some subset of objects. It follows that , which means that . Applying the change of base formula gives us .
Final answer
n