Accedmychevron_right11thchevron_rightmathchevron_rightMathematical Inductions And Binomial Theoremchevron_rightExercise 8.1

Questions

  1. Question 1

    Use mathematical induction to prove the following formulae for every positive integer nn.

    (i)

    logxn=nlogx\log x^n = n \log x, where xx is positive

    (ii)

    2+5+8++(3n1)=n2(3n+1)2 + 5 + 8 + \cdots + (3n-1) = \dfrac{n}{2}(3n+1)

    (iii)

    2+(2+5)+(2+5+8)++n2(3n+1)=n4(n+1)22 + (2+5) + (2+5+8) + \cdots + \dfrac{n}{2}(3n+1) = \dfrac{n}{4}(n+1)^2

    (iv)

    2+6+18++2×3n1=3n12 + 6 + 18 + \cdots + 2 \times 3^{n-1} = 3^n - 1

    (v)

    1×3+2×5+3×7++n×(2n+1)=n(n+1)(4n+5)61\times 3 + 2\times 5 + 3\times 7 + \cdots + n\times(2n+1) = \dfrac{n(n+1)(4n+5)}{6}

    (vi)

    11×2+12×3+13×4++1n(n+1)=11n+1\dfrac{1}{1\times 2} + \dfrac{1}{2\times 3} + \dfrac{1}{3\times 4} + \cdots + \dfrac{1}{n(n+1)} = 1 - \dfrac{1}{n+1}

    (vii)

    r+r2+r3++rn=r(1rn)1rr + r^2 + r^3 + \cdots + r^n = \dfrac{r(1-r^n)}{1-r}, (r1)(r \neq 1)

    (viii)

    a+(a+d)+(a+2d)++[a+(n1)d]=n2[2a+(n1)d]a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] = \dfrac{n}{2}[2a+(n-1)d]

    (ix)

    an=a1+(n1)da_n = a_1 + (n-1)d when a1,a1+d,a1+2d,a_1, a_1+d, a_1+2d, \ldots form an A.P.

    (x)

    an=a1rn1a_n = a_1 r^{n-1} when a1,a1r,a1r2,a_1, a_1 r, a_1 r^2, \ldots form a G.P.

    (xi)

    (33)+(43)+(53)++(n+23)=(n+34)\dbinom{3}{3} + \dbinom{4}{3} + \dbinom{5}{3} + \cdots + \dbinom{n+2}{3} = \dbinom{n+3}{4}

    (xii)

    The sum of first nn odd natural numbers is n2n^2.

  2. Question 3

    Prove that k=1nrk=rn+11r1\sum_{k=1}^{n} r^k = \dfrac{r^{n+1}-1}{r-1}, whenever nn is a positive integer.

  3. Question 4

    xyx - y is a factor of xnynx^n - y^n for all positive integral values of nn, (xy)(x \neq y).

  4. Question 5

    n!>2n1n! > 2^n - 1 for integral values of n4n \ge 4.

  5. Question 6

    4n>3n+2n14^n > 3^n + 2^{n-1} for integral values of n2n \ge 2.

  6. Question 7

    1+nx(1+x)n1 + nx \le (1+x)^n for n2n \ge 2 and x>1x > -1.

  7. Question 8

    Aliza invests Rs. 1,000,000 in a business that promises a 6% return compounded annually. Prove by mathematical induction that the amount of money after nn years is 1,000,000(1.06)n1,000,000(1.06)^n.

  8. Question 9

    A bank offers an investment with an annual interest rate rr. If PP rupees are invested, the amount after nn years is given by: A(n)=P(1+r)nA(n) = P(1+r)^n. Prove by induction that this formula holds for all n0n \ge 0.

  9. Question 10

    Sikander saves Rs. 500 in the first month and increases his savings by Rs. 500 every subsequent month. Using mathematical induction, determine whether his total savings will reach at least Rs. 12,000 after 24 months.

  10. Question 11

    Prove by mathematical induction that if Ali takes a loan of Rs. 2,000,000 and pays Rs. 50,000 at the end of each year, the remaining balance after nn years is Rn=2,000,00050,000nR_n = 2,000,000 - 50,000n.

  11. Question 12

    If Salman starts savings with Rs. 5,000 and saves an additional Rs. 1,000 at the end of every month, derive a formula S(n)S(n) for his total savings after nn months. Prove the correctness of your formula using mathematical induction.