Answer:
T⁵ + 5T⁴ + 10T³ + 21T² + 2T - 2.
Explanation:
standard form of the expression 11t+2t²-3+T to the fifth:
We start with the expression:
11t + 2t² - 3 + T⁵.
We can use the binomial theorem to expand (T + 1)⁵.
The formula for the binomial theorem is:
(a + b)ⁿ = ∑(k=0 to n) [n choose k] a^(n-k) b^k
where [n choose k] is the binomial coefficient, given by:
[n choose k] = n! / (k! (n-k)!)
Applying this formula with a = T and b = 1, we get:
(T + 1)⁵ = ∑(k=0 to 5) [5 choose k] T^(5-k) 1^k
= [5 choose 0] T⁵ + [5 choose 1] T⁴ + [5 choose 2] T³ + [5 choose 3] T² + [5 choose 4] T + [5 choose 5] 1
= T⁵ + 5T⁴ + 10T³ + 10T² + 5T + 1
We can substitute this expansion into the original expression:
11t + 2t² - 3 + T⁵ = 11t + 2t² - 3 + (T⁵ + 5T⁴ + 10T³ + 10T² + 5T + 1)
= T⁵ + 5T⁴ + 10T³ + 10T² + (5T + 11t) - 2
Finally, we write the terms in descending order of their degree, and combine like terms:
T⁵ + 5T⁴ + 10T³ + 21T² + 2T - 2
Therefore, the standard form of the expression 11t+2t²-3+T to the fifth is
T⁵ + 5T⁴ + 10T³ + 21T² + 2T - 2.