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.