The number of integral terms in the binomial expansion of \[ \left(11^{\frac{1}{2}} + 17^{\frac{1}{8}}\right)^{1024} \] is:

Question

If $a + b = 5$ and $ab = 6$, find $a^2 + b^2$:

Answer 1

To solve the problem of finding the number of integral terms in the binomial expansion of \left(11^{\frac{1}{2}} + 17^{\frac{1}{8}}\right)^{1024} , we need to consider the general term in the expansion and determine when it becomes an integer.

The general term of a binomial expansion (a + b)^n is given by:

T_{k+1} = \binom{n}{k} \cdot a^{n-k} \cdot b^k

Here, a = 11^{\frac{1}{2}} , b = 17^{\frac{1}{8}} , and n = 1024 .

The general term T_{k+1} becomes:

T_{k+1} = \binom{1024}{k} \cdot (11^{\frac{1}{2}})^{1024-k} \cdot (17^{\frac{1}{8}})^k = \binom{1024}{k} \cdot 11^{\frac{1024-k}{2}} \cdot 17^{\frac{k}{8}}

For T_{k+1} to be an integer, 11^{\frac{1024-k}{2}} \cdot 17^{\frac{k}{8}} must be an integer. This requires both \frac{1024-k}{2} and \frac{k}{8} to be integers.

Thus, \frac{1024-k}{2} = m and \frac{k}{8} = n where m and n are integers.

This gives:

1024 - k = 2m \Rightarrow k = 1024 - 2m
k = 8n

Equating the two expressions for k :

8n = 1024 - 2m

Simplifying, we have:

2m + 8n = 1024 \\ m + 4n = 512

This is a linear Diophantine equation. The solutions exist for all integer values of n such that m \geq 0 .

From the equation m + 4n = 512 , we get:

m = 512 - 4n

So, m \geq 0 \Rightarrow 512 - 4n \geq 0 \Rightarrow n \leq 128 .

Thus, n ranges from 0 to 128 .

For each n , there exists a corresponding k \: k = 8n \ in the range of [0, 1024] .

This gives us 129 possible values for n and hence 129 integral terms.

Therefore, the number of integral terms in the given binomial expansion is 129 .

Answer 2