Sum of Binomial Coefficients

Putting x = 1 in the expansion (1+x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² +...+ ⁿC_x xⁿ, we get,

2ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ +...+ ⁿC_n.

We kept x = 1, and got the desired result i.e. ∑ⁿ_r=0  C_r = 2ⁿ.

Note:  This one is very simple illustration of how we put some value of x and get the solution of the problem. It is very important how judiciously you exploit this property of binomial expansion.

Illustration:

Find the value of C₀ + C₂ + C₄ +... in the expansion of (1+x)ⁿ.

Solution:

We have,

        (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² +... ⁿC_n xⁿ.

Now put x = -x; (1-x)ⁿ = ⁿC₀ - ⁿC₁ x + ⁿC₂ x² -...+ (-1)^{n n}C_n xⁿ.

        Now, adding both expansions, we get,

        (1 + x)ⁿ + (1-x)ⁿ = 2[ⁿC₀ + ⁿC₂ x² + ⁿC₄ x⁴ +.........]

Put    x = 1

        =>     (2ⁿ+0)/2 = C₀ + C₂ + C₄ +......

        or     C₀ + C₂ + C₄ +......= 2^n-1

We have found the sum of binomial coefficients. But if these coefficients are multiplied by some factors can we find the sum for such expressions?

Yes, we can often find it by creatively applying what we have learnt. Let us see how differentiation and integration are useful in these situations.

Suppose we want to calculate the value of

        ∑ⁿ_r=0  C_r i.e. 0.C₀ + 1 C₁ + 2C₂ +...+ n ⁿC_n

If we take a close look to the sum to be found, we find that coefficients are multiplied with respective powers of x.

 If we want to multiply the coefficient of x by its power differentiation is of help. Hence differentiate both sides of

        (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ⁿC₃ x³ +...+ ⁿC_n xⁿ, with respect to x we get

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

        Put x = 1, we get, n 2^n-1 = + 1 C₁ + 2 C₂ +...+ n C_n.

Or,      ∑ⁿ_r=0     r  C_r = n 2^n-1, which is the answer.

        Now, think how to find the following sum

        C₀/1 + C₁/2 + C₂/3 +......+ C_n/n+1 = ?

        In this sum coefficients are divided by the respective power of x + 1. This expression can be achieved by Integrating the expansion of (1 + x)ⁿ under proper limits.

 Important:  It has said earlier in this chapter that we should use and exploit the property that x can take any value in the expansion of (1 + x)ⁿ.

         Now, let us try to find the value of

        C₀ - C₂ + C₄-C₆ +.........

        Analysing the above, expression, we find that C₀, C₂, C₄ are all coefficients of even powers of x.

        Had it been like C₀ + C₂ + C₄ +...... we could have evaluated simply by the method described earlier or had it been like C₀-C₁ + C₂-C₃ +... we could have put x = -1 in the expansion of (1 + x)ⁿ and find the sum.

But here the case is different. The expression consists of coefficients of only even powers. In such cases when there is alternative sign charge for the coefficient, of same nature of powers (even or odd) use of I (iota) comes to our rescue.

                (1 + x)ⁿ = C₀ + C₁ x + C₂ x² +......

                (1 - x)ⁿ = C₀ - C₁ x + C₂ x² +......

By adding   (1 + x)ⁿ + (1 - x)ⁿ = 2[C₀ + C₂ x² + C₄ x⁴ +......].

        Clearly use of √-1 (iota) will generate the given expression.

                ((1+i)ⁿ + (1-i)ⁿ)/2 ] [C₀ - C₂ + C₄ + ......]

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