Evaluate sum_k=1^n (k2 + k).

11. Evaluate _k=1^n (k^2 + k). Step 1: Separate the summation into two parts. _k=1^n (k^2 + k) = _k=1^n k^2 + _k=1^n k Step 2: Apply the standard summation formulas. The formula for the sum of the first n squares is _k=1^n k^2 = (n(n+1)(2n+1))/(6). The formula for the sum of the first n natural numbers is _k=1^n k = (n(n+1))/(2). Substitute these formulas into the expression: _k=1^n (k^2 + k) = (n(n+1)(2n+1))/(6) + (n(n+1))/(2) Step 3: Find a common denominator and combine the terms. The common denominator is 6. _k=1^n (k^2 + k) = (n(n+1)(2n+1))/(6) + (3n(n+1))/(6) _k=1^n (k^2 + k) = (n(n+1)(2n+1) + 3n(n+1))/(6) Step 4: Factor out common terms and simplify. Factor out n(n+1) from the numerator: _k=1^n (k^2 + k) = (n(n+1)[(2n+1) + 3])/(6) _k=1^n (k^2 + k) = (n(n+1)(2n+4))/(6) Factor out 2 from (2n+4): _k=1^n (k^2 + k) = (n(n+1)2(n+2))/(6) _k=1^n (k^2 + k) = (2n(n+1)(n+2))/(6) _k=1^n (k^2 + k) = (n(n+1)(n+2))/(3) The evaluated sum is (n(n+1)(n+2))/(3). 12. The 13th and 16th terms of an A.P. are 80 and 98 respectively. Find the first term, the common difference and the last term if there are 20 terms. Step 1: Set up equations using the formula for the n-th term of an A.P., a_n = a + (n-1)d. Given a_13 = 80: a + (13-1)d = 80 a + 12d = 80 (Equation 1) Given a_16 = 98: a + (16-1)d = 98 a + 15d = 98 (Equation 2) Step 2: Solve the system of equations for a and d. Subtract Equation 1 from Equation 2: (a + 15d) - (a + 12d) = 98 - 80 3d = 18 d = (18)/(3) d = 6 Substitute d=6 into Equation 1: a + 12(6) = 80 a + 72 = 80 a = 80 - 72 a = 8 Step 3: State the first term and common difference. The first term is 8. The common difference is 6. Step 4: Find the last term if there are 20 terms. The last term is a_20. Using the formula a_n = a + (n-1)d: a_20 = 8 + (20-1)6 a_20 = 8 + (19)6 a_20 = 8 + 114 a_20 = 122 The last term is 122.