The value of E=(1+17)1+1721+173…1+1719(1+19)1+1921+193…1+1917 is

We have(1+k)1+k21+k3…1+kn=(1+k)(2+k)(3+k)…(n+k)(2)(3)…(n)=(n+k)!k!n!=n+kCkThus, both the numerator and the denominator of E equals  36C17=36C19 â‹…∴ E=1.