Some Important Results

1.     Differentiating (1+x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ of both sides we have,

        n(1 + x)^n-1 = C₁ + 2C₂x + 3C₃x² +...+ ⁿC_nx^n-1.                   ... (E)

        Put x = 1 in (E) so that n2ⁿ⁺¹ = C₁ + 2C₂ + 3C₃ + ...+ nC_n.

        Put x = -1 in (E) so that 0 = C₁ - 2C₂ +...+ (-1)^{n-1 n}C_n.

        Differentiating (E) again and again we will have different results.

2.     Integrating (1 + x)ⁿ, we have,

        ((1+x)ⁿ⁺¹)/(n+1) + C = C₀x +  C₁x²/2  + C₂x³/3 +.....+C_nxⁿ⁺¹/(n+1)        (where C is a constant)

        For x = 0, we get C = -1/(n+1) .

        Therefore ((1+x)ⁿ⁺¹ - 1)/(n+1)  = C₀x +  C₁x²/2  + C₂x³/3 +.....+C_nxⁿ⁺¹/(n+1)  ..... (F)

        Put x = 1 in (F) and get

         (2ⁿ⁺¹ - 1)/n+1= C₀ + C₁/2 +...C_n/n+1 .

        Put x = -1 in (F) and get, 1/n+1 = C0 - C₁/2 + C₂/3 -  ......

        Put x = 2 in (F) and get, (3ⁿ⁺¹-1)/n+1 = 2 C₀ + 2²/2 C₁ + 2²/3 C₂ +...+ 2ⁿ⁺¹/n+1  = C_n.

Problems Related to Series of Binomial Coefficients in Which Each Term is a Product of an Integer and a Binomial Coefficient, i.e. In the Form k.ⁿC_r. 

Illustration:

       If (1+x)n =x^r then prove that C₁ + 2C₂ + 3C₃ +...+ nC_n = n2^n-1.

Solution:

          Method (i) : r^th term of the given series

        r^th term of the given series, t_r = nC_r

        => t_r = r × n/r × ^n-1C_r-1 = n × ^n-1C_r-1          (because ⁿC_r =n/r .^n-1C_r-1)

        Sum of the series =

        Put x = 1 in the expansion of (1 + x)^n-1, so that

        (^n-1C₀ + ^n-1C₁ +...+  ^n-1C_n-1) = 2n-1

        => = n.2n-1.

          Method (ii) : By Calculus

        We have (1 + x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ.                       ... (1)

        Differentiating (1) w.r.t. x, we get

        n(1 + x)^n-1 = C₁ + 2C₂x + 3C₃ x² +...+ n C_nx^n-1.                     ... (2)

        Putting x = 1 in (2), we have, n 2^n-1 = C₁ + 2C₂ +....+ nC_n.         ... (3)

Illustration:

If (1+ x)ⁿ =  x^r then prove that C₀ + 2.C₁ + 3.C₂+...+(n+1)C_n=^2n-1(n+2).

Solution:

Method (i): r^th term of the given series

        rth term of the given series

        t_r = ⁿC_r-1 = [(r-1) + 1]. ⁿC_r-1

        = (r-1) ⁿC_r-1 + ⁿC_r-1 = n. ^n-1C_r-2 + ⁿC_r-1 (because ⁿC_r-1 =n/(r-1) . ^n-1C_r-2)

        Sum of the series =

        = n[^n-1C₀ + ^n-1C₁ +...+ ^n-1C_n-1]+[ⁿC₀ + ⁿC₁ +...+ ⁿC_n] = n.2^n-1 + 2n
        = 2^n-1 (n+2).

Method (ii) by Calculus.

        We have (1 + x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ.                        ... (1)

        Multiplying (1) with x, we get

        x(1+x)ⁿ = C₀x + C₁x² + C₂x³ +...+ C_nxⁿ⁺¹.                                ... (2)

        Differentiating (2) w.r.t. x, we have

        (1 + x)ⁿ + n(1 + x)^n-1 x = C₀x + 2C₁x² +...+ (n+1)C_nxⁿ                ... (3)

        Putting x = 1 in (3), we get

        2n + n.2^n-1 = C₀ + 2C₁ + 3C₂ +...+ (n+1)C_n

        => C₀ + 2C₁ + 3C₂ +...+ (n+1)C_n = 2^n-1 (n+2). 

Problems Related to Series of Binomial Coefficient in Which Each Term is Binomial Coefficient divided by an Integer, i.e. in the Form of ⁿC_r/k.

Illustration:

        If (1+x)ⁿ =

Solution:

Method (i) : r^th term of the given series

Method (ii): By Calculus

       (1+x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ                                        ... (1)

       Integrating both the sides of (1) w.r.t. x between the limits 0 to x, we get

           ... (2)

      Substituting x = 1 in (2), we get 2ⁿ⁺¹/n+1 = C_{0 +} c₁/2 + c₂/3 +_.....+ c_n/n+1.

Illustration:

       If (1+x)n  = x^r ,show that

Method I : r^th term of the given series T_r =

Method II : (By Calculus)

       (1 + x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ

        => x(1+x)ⁿ = C₀x + C₁x² + C₂x³ +...+ C_nxⁿ⁺¹                      ... (1)

        Integrating both the sides of (1) with respect to x

        Put x = 0, => k = 1/((n+1)(n+2))

        Put x = 1,

        .

Problem Related to Series of Binomial Coefficients in Which Each Term is a Product of two Binomial Coefficients.

(a) If sum of lower suffices of binomial expansion in each term is the same

        i.e. ⁿC₀ ⁿC_n + ⁿC₁ ⁿC_n-1 + ⁿC_n-2 +...+ ⁿC_n ⁿC₀

        i.e. 0 + n = 1 + (n-1) = 2 + (n-2) = n + 0.

Then the series represents the coefficients of xn in the multiplication of the  following two series

       (1+x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ

       and (1+x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ.

Illustration:

       Prove that C₀C_r + C₁C_r+1 + C₂C_r+2 +...+ C_n-r C_n = (2n)!/((n-r)!(n+r)!)

Solution:

        We have,

        C₀ + C₁x + C₂x² + ... + C_nxⁿ = (1+x)ⁿ                                      ... (1)

        Also C₀xⁿ + C₂x^n-2 +...+ C_n = (x+1)ⁿ                                        ... (2)

        Multiplying (1) and (2), we get

        (C₀ + C₁x² +...+ C_nxⁿ)(C₀xⁿ + C₁x^n-2 + C₂x^n-2 +...+ C_n) = (1+x)²ⁿ ... (3)

        Equating coefficient of x^n-r from both sides of (3), we get

        C₀C_r + C₁C_r+1 + C₂C_r+2 +...+ C_n-rC_n = 2_nC_n-r = (2n)/((n-r)!(n+r)!) .

Illustration:

        Prove that C_o² + C₁² +...+ C_n² = (2n)!/n!n! .

Solution:

        Since

        (1 + x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ,                                     ... (1)

        (x + 1)ⁿ = C₀xⁿ + C₁x^n-2 +...+ C_n,                                            ... (2)

        (C₀ + C₁x + C₂x² +...+ C_nxⁿ)(C₀xⁿ + C₁x^n-1 + C₂x^n-2 +...+ C_n)=(1+x)²ⁿ.

        Equating coefficient of xⁿ, we get

        C₀² + C₁² + C₂² +...+ C_n² = ²ⁿC_n = (2n)!/n!n! .

Illustration:

        If (1+x)ⁿ = , then prove that

        ^mC_r ⁿC₀ + ^mC_r-1 ⁿC₁ + ^mC_r-2 ⁿC₂ +...+ ^mC₁ ⁿC_r-1 + ^mC₀ ⁿC_r = ^m+nC_r

        where m, n, r are positive integers and r < m and r < n.

Solution:

        (1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² +...+ ⁿC_rx^r +...+ ⁿC_nxⁿ                      ... (1)

        and also

        (1+x)^m = ^mC₀ + ^mC₁x + ^mC₂x² +...+ ^mC_rx^r +...+ ^mC_mx^m                ... (2)

        Multiplying (1) and (2), we get

        (ⁿC₀ + ⁿC₁x + ⁿC₂x² +...+ ⁿC_rx^r +...+ ⁿC_nxⁿ)x

        (^mC₀ + ^mC₁x + ^mC₂x² +...+ ^mC_rx^r +...+ ^mC_mx^m) = (1+x)^m+n

        = ^m+nC₀ + ^m+2C₁x + ^m+nC₂x² +...+ ^m+nC_rx^r +...+ ^m+nC_m+nx^m+n

        Equating the coefficient of x^r, we get

        ^mC_r ⁿC₀ + ^mC_r-1 ⁿC₁ + ^mC_r-2 ⁿC₂ +...+ ^mC₁ ⁿC_r-1 + ^mC₀ ⁿC_r = ^m+nC_r

(b)    If one series has constant lower suffices and other has varying lower suffices

Illustration:

        Prove that ⁿC₀.²ⁿC_n - ⁿC₁^2n-2C_n + ⁿC₂/^2n-4C_n -...= 2_n.

Solution:

        ⁿC₀.²ⁿC_n - ⁿC₁^2n-2C_n + ⁿC₂/^2n-4C_n -...

        = coefficient of xⁿ in[ⁿC₀(1+x)²ⁿ - ⁿC₁(1+x)^2n-2+ⁿC₂(1+x)^2n-4 - ...]

        = coefficient of xⁿ in

        [ⁿC₀((1+x)²)ⁿ - ⁿC₁((1+x)²)^n-1 + ⁿC₂((1+x)²)^n-2 - ...]

        = coefficient of xⁿ in [(1+x)² - 1]ⁿ

        = coefficient of xⁿ in (2x+x²)ⁿ = co-efficient of xⁿ in xⁿ (2+x)ⁿ = 2ⁿ.

At askIITians we provide you free study material on Set Theory and Functions and you get all the professional help needed to get through IIT JEE and AIEEE easily. AskIITians also provides live online IIT JEE preparation and coaching where you can attend our live online classes from your home!