Let f(n) be the base-10 logarithm of the sum of the elements of the nth row in Pascal's triangle. Express f(n)log102 in terms of n. Recall that Pascal's triangle begins
1 -> n=0
1 1 -> n=0
1 2 1 -> n=0
1 3 3 1 -> n=0
1 4 6 4 1 -> n=0
... ...
the nth row of Pascal's triangle is given by{(nk):k=0,n}
f(n)=log10(n∑k=0(nk))=log10(2n)=nlog10(2)f(n)log10(2)=n
+88 
