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03.14 · Algebra

Binomial Theorem

Elite lesson11 formulas10 questions

On the syllabus of

Elite lesson38 minHinglish

Formulas

At a glance 11 formulas

  1. 1General termTr+1=(nr) a n−r b rT_{r+1} = \binom{n}{r}\,a^{\,n-r}\,b^{\,r}in (a+b)n, r=0,1,…,n (n+1 terms)\text{in } (a + b)^n,\ r = 0, 1, \dots, n \ (n + 1 \text{ terms})
  2. 2Middle termn even: Tn2+1n \text{ even: } T_{\frac{n}{2} + 1}n odd: Tn+12 and Tn+32n \text{ odd: } T_{\frac{n+1}{2}} \text{ and } T_{\frac{n+3}{2}}
  3. 3Symmetry(nr)=(nn−r)\binom{n}{r} = \binom{n}{n - r}the largest (nr) is the middle one\text{the largest } \binom{n}{r} \text{ is the middle one}
  4. 4Term independent of xWrite the power of xx in Tr+1T_{r+1} and set it to 0 (or to kk for the xkx^k term). If no whole-number rr fits, there is no such term.
  5. 5Sum of coefficientssum of coefficients of f(x)=f(1)\text{sum of coefficients of } f(x) = f(1)(n0)+(n1)+⋯+(nn)=2n\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n
  6. 6Even and odd powerseven powers of f(x)=f(1)+f(−1)2\text{even powers of } f(x) = \dfrac{f(1) + f(-1)}{2}odd powers=f(1)−f(−1)2, constant=f(0)\text{odd powers} = \dfrac{f(1) - f(-1)}{2},\ \text{constant} = f(0)
  7. 7Weighted sum∑rr(nr)=n 2 n−1\textstyle\sum_{r} r\binom{n}{r} = n\,2^{\,n-1}since r(nr)=n(n−1r−1)\text{since } r\binom{n}{r} = n\binom{n-1}{r-1}
  8. 8Neighbouring coefficients(nr)÷(nr−1)=n−r+1r\binom{n}{r} \div \binom{n}{r-1} = \dfrac{n - r + 1}{r}
  9. 9Combine before expandingMultiply brackets first when they pair up: (1+x)n(1−x)n=(1−x2)n(1 + x)^n(1 - x)^n = (1 - x^2)^n, so only even powers of xx survive.
  10. 10Remainders(1+m)n=1+nm+(multiple of m2)(1 + m)^n = 1 + nm + (\text{multiple of } m^2)so (1+m)n÷m2 leaves what 1+nm leaves\text{so } (1 + m)^n \div m^2 \text{ leaves what } 1 + nm \text{ leaves}
  11. 11Rational termsIn (Ap+Bq)n(\sqrt[p]{A} + \sqrt[q]{B})^n, with no perfect powers inside the roots, Tr+1T_{r+1} is rational exactly when pp divides n−rn - r and qq divides rr.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    What is the coefficient of x2x^2 in the expansion of (1+2x)6(1 + 2x)^6?

    Correct answer60

    Why

    General term: Tr+1=(6r)(2x)rT_{r+1} = \binom{6}{r}(2x)^r, and x2x^2 needs r=2r = 2.

    The coefficient is (62)×22=15×4\binom{6}{2} \times 2^2 = 15 \times 4 (leaving out the 222^2 gives the trap 15).

    So the coefficient of x2x^2 is 60.

  2. Q2Easy

    What is the sum of all the coefficients in the expansion of (2x−3y)7(2x - 3y)^7?

    Correct answer−1-1

    Why

    Sum of coefficients: put every variable equal to 1.

    That gives (2−3)7=(−1)7(2 - 3)^7 = (-1)^7; dropping the minus sign would give the trap 57=781255^7 = 78125.

    So the coefficients add up to −1-1.

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