a) Profit is the difference between revenue and cost. The revenue from the sale of radios will be ...
(revenue in year n) = (units sold in year n)*(price per radio in year n)
r(n) = (450 +50n)*(20 +n^2/45) = (10/9)n^3 +10 n^2 +1000n +9000
The cost for the production of all radios sold is given. Profit is then ...
(profit in year n) = (revenue in year n) - (cost in year n)
(profit in year n) = (10/9)n^3 +10n^2 +1000n +9000 -(1000 +10n^2)
(profit in year n) = (10/9)n^3 +1000n +8000
P(n) = (10/9)(n^3 +900n +7200)
b) The sum of numbers 1 to n is given by n(n+1)/2. The sum of numbers 1 to n^3 is given by (n(n+1)/2)^2. The sum of n constants is n*(the constant). Thus the sum T(n) of the terms of P(n) will be the sum of these polynomials multiplied by appropriate factors. It can be written in the form shown in the problem statement.
c) T(20) is computed to be 419,000, so is a little short of predicting the actual value of the contract.