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Evaluate the following sums: a) sumk=120 (3k+2) b) sumk=14 (k2+k)

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11. Evaluate the following sums:

Part a· 3 steps

∑k=120(3k+2)\sum_{k=1}^{20} (3k+2)

  1. Use the linearity property of summation to split the sum.

    ∑k=120(3k+2)=∑k=1203k+∑k=1202\sum_{k=1}^{20} (3k+2) = \sum_{k=1}^{20} 3k + \sum_{k=1}^{20} 2

  2. Factor out the constant from the first term and apply the sum formulas ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2} and ∑k=1nc=nc\sum_{k=1}^{n} c = nc.

    Here n=20n=20.

    =3∑k=120k+∑k=1202= 3 \sum_{k=1}^{20} k + \sum_{k=1}^{20} 2 =3(20(20+1)2)+(2×20)= 3 \left( \frac{20(20+1)}{2} \right) + (2 \times 20)

  3. Calculate the values.

    =3(20×212)+40= 3 \left( \frac{20 \times 21}{2} \right) + 40 =3(10×21)+40= 3 (10 \times 21) + 40 =3(210)+40= 3(210) + 40 =630+40= 630 + 40 =670= 670

    The sum is 670\boxed{670}.

Part b· 7 steps

∑k=14(k2+k)\sum_{k=1}^{4} (k^2+k)

  1. Use the linearity property of summation to split the sum.

    ∑k=14(k2+k)=∑k=14k2+∑k=14k\sum_{k=1}^{4} (k^2+k) = \sum_{k=1}^{4} k^2 + \sum_{k=1}^{4} k

  2. Apply the sum formulas ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} and ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}.

    Here n=4n=4.

    =4(4+1)(2×4+1)6+4(4+1)2= \frac{4(4+1)(2 \times 4+1)}{6} + \frac{4(4+1)}{2}

  3. Calculate the values.

    =4(5)(8+1)6+4(5)2= \frac{4(5)(8+1)}{6} + \frac{4(5)}{2} =4×5×96+202= \frac{4 \times 5 \times 9}{6} + \frac{20}{2} =1806+10= \frac{180}{6} + 10 =30+10= 30 + 10 =40= 40

    The sum is 40\boxed{40}.

    12. The 13th13^{th} and 16th16^{th} 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.

  4. Write down the given information using the formula for the nthn^{th} term of an A.P., an=a1+(n−1)da_n = a_1 + (n-1)d.

    For the 13th13^{th} term:

    a13=a1+(13−1)d  ⟹  a1+12d=80(Equation 1)a_{13} = a_1 + (13-1)d \implies a_1 + 12d = 80 \quad \text{(Equation 1)}

    For the 16th16^{th} term:

    a16=a1+(16−1)d  ⟹  a1+15d=98(Equation 2)a_{16} = a_1 + (16-1)d \implies a_1 + 15d = 98 \quad \text{(Equation 2)}

  5. Solve the system of equations to find the common difference dd.

    Subtract Equation 1 from Equation 2:

    (a1+15d)−(a1+12d)=98−80(a_1 + 15d) - (a_1 + 12d) = 98 - 80 3d=183d = 18 d=183d = \frac{18}{3} d=6d = 6

    The common difference is 6\boxed{6}.

  6. Substitute the value of dd into Equation 1 to find the first term a1a_1.

    a1+12(6)=80a_1 + 12(6) = 80 a1+72=80a_1 + 72 = 80 a1=80−72a_1 = 80 - 72 a1=8a_1 = 8

    The first term is 8\boxed{8}.

  7. Find the last term, which is the 20th20^{th} term (a20a_{20}), since there are 20 terms.

    Using the formula an=a1+(n−1)da_n = a_1 + (n-1)d with n=20n=20, a1=8a_1=8, and d=6d=6:

    a20=8+(20−1)6a_{20} = 8 + (20-1)6 a20=8+(19)6a_{20} = 8 + (19)6 a20=8+114a_{20} = 8 + 114 a20=122a_{20} = 122

    The last term is 122\boxed{122}.

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