1.1 Evaluating expressions by Differentiation
Often, we would need to call up Calculus to compute Binomial Expressions and sums. We’ll represent\(C(n,r)\) as \(C_r\) from here on. Let’s consider a small example:
Ex -1:
Show that \(C_1+2C_2+3C_3+…+nC_n=n2^{n-1}\)
Solution:
If the positive integers appear as the product of Binomial coefficients, then that’s a hint that the problem may be solvable by applying differentiation concepts.
Consider \((1+x)^n=c_0+C_1x+C_2x^2+…+C_nx^n\).
Differentiating both sides with respect to \(x\) gives
\(n(1+x)^{n-1}=0+C_1+2C_2x+3C_3x^2+…nC_nx^{n-1}\).
Putting \(x=1\), we get \(C_1+2C_2+3C_3+…+nC_n=n2^{n-1}\).
In general, to compute such a series, we may need to look at the magnitude of the integer multiplier of the last coefficient of the series. If the greatest linear factor of that integer is \(m\), then divide \(m\) by \(n\). Let’s call \(q\) as the quotient and \(r\) as the remainder of this division. Then, replace \(x\rightarrow x^q\) in the series, and multiply both sides of the expressions by \(x^r\). However, if the multiplies are products of 2 integers, or squares and cubes of integers, then we may need to differentiate multiple times.
Let’s look at another example:
Ex -2:
Compute \(1^2C_1+2^2C_2+…+n^2C_n\)
Solution:
The linear factors of the last term are \(n,n\), and \(n=1\times n+0\).
Thus, considering \((1+x)^n=C_0+C_1x+C_2x^2+…+C_nx^n\), and differentiating both sides w.r.t \(x\), we have
\(n(1+x)^{n-1}=C_1+2C_2x+3C_3x^2+…+nC_nx^{n-1}\).
Again, we see the last term as \(n\), thus dividing the same by \(n-1\) (which is the power of \(x\) in the last term), we have \(n=(n-1)\times 1 +1\). We therefore multiply both sides by \(x\) to have
\(nx(1+x)^{n-1}=C_1x+2C_2x^2+…+nC_nx^n\).
Differentiate w.r.t \(x\) and putting \(x=1\), we get
\(1^2C_1+2^2C_2+…n^2C_n=n(n+1)2^{n-2}\).
1.2 Evaluating expressions by Integration
Integration is preferred over differentiation, if the integers appear as denominators of the binomial coefficients. Based on the signs and the binomial coefficients appearing in the expressions, we may need to integrate between different limits. Some rules of the thumb may be:
- If the expression contains \(C_0,C_1,C_2,…\) all with positive signs, then try integrating between 0 and 1
- If the sum contains alternate + and – signs, then try integrating between -1 and 0.
- If the sum contains alternate odd coefficients \(C_0,C_2,C_4,…\), then try integrating from -1 to +1
- If the sum contains alternate even coefficients \(C_1,C_3,C_5,..\), try subtracting (ii) from (i) and then divide by 2.
- If the denominator of binomial coefficients is the product of 2 integers, then 2 times integration maybe required.
Let’s analyse with an example:
Ex-3:
Show that \(\sum\limits_{r=0}^n(-1)^r\frac{C_r}{r+1}=\frac{1}{n+1}\)
Solution:
Integrating \((1+x)^n\) from -1 to 0 we have
\(\int_{-1}^0(1+x)^ndx = \int_{-1}^0 (\sum\limits_{r=0}^nC_rx^r)dx\Rightarrow \left[ \frac{(1+x)^n}{n+1}\right]_{-1}^0=\left[ \sum\limits_{r=0}^nC_r\frac{x^{r+1}}{r+1}\right]_{-1}^0\)This gives \(\frac{1}{n+1}=\sum\limits_{r=0}^n(-1)^r\frac{C_r}{r+1}\)