Let a=1+ 2C23!+ 3C24!+ 4C25!+ , b=1+ 1C0+1C11!+ 2C0+2C1+2C22! + 3C0+3C1+3C2+3C33!+ . Then 2ba2 is equal to .
Answer
8
Solution
Since nC0 + nC1 + + nCn = 2n, we get b = 1 + 21! + 222! + 233! + = e2.
For a, expand e1 + x in powers of (1 + x), remove 1 and divide by (1 + x):
e1 + x - 11 + x = 1 + (1 + x)2! + (1 + x)23! + (1 + x)34! +
The coefficient of x2 on the right is 2C23! + 3C24! + = a - 1.
On the left, (e - 1 + ex + e2x2 + ) (1 - x + x2 - ) has x2-coefficient (e - 1) - e + e2 = e2 - 1.
Equating the two, a = e2.
2ba2 = 2e2e2/4 = 8
Answer: 8