Understanding the Binomial Expansion and Coefficients

The problem asks us to find the sum of terms involving the coefficients \(T_k\) from the binomial expansion of \((1+x)^n\). The expansion is given as:

\((1 + x)^n = 1 + T_1x + T_2x^2 + T_3x^3 + \dots + T_nx^n\)

We know the standard binomial expansion is given by the Binomial Theorem:

\((1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n\)

Comparing these two forms, we can identify the coefficients \(T_k\) for \(k \ge 1\). Note that the constant term in the given expansion is 1, which matches \(\binom{n}{0} = 1\). So, for \(k \ge 1\), the coefficient \(T_k\) of \(x^k\) is equal to the binomial coefficient \(\binom{n}{k}\). Specifically:

• \(T_1 = \binom{n}{1}\)
• \(T_2 = \binom{n}{2}\)
• \(T_3 = \binom{n}{3}\)
• ...
• \(T_n = \binom{n}{n}\)

The sum we need to evaluate is \(S = T_1 + 2T_2 + 3T_3 + \dots + nT_n\).

Substituting the binomial coefficients for \(T_k\), this sum is:

\(S = 1 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + 3 \cdot \binom{n}{3} + \dots + n \cdot \binom{n}{n}\)

We can write this sum using summation notation:

\(S = \sum_{k=1}^n k \binom{n}{k}\)

Method 1: Using Combinatorial Identity for Binomial Coefficient Sums

We can use a known combinatorial identity that relates the term \(k \binom{n}{k}\) to another binomial coefficient. The identity is:

\(k \binom{n}{k} = n \binom{n-1}{k-1}\) for \(k \ge 1\)

Let's briefly show why this identity is true:

\(k \binom{n}{k} = k \frac{n!}{k!(n-k)!}\)

\(= \frac{n!}{(k-1)!(n-k)!}\)

\(= n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\)

\(= n \binom{n-1}{k-1}\)

Now substitute this identity into the sum \(S\):

\(S = \sum_{k=1}^n k \binom{n}{k} = \sum_{k=1}^n n \binom{n-1}{k-1}\)

Since \(n\) is a constant with respect to the summation variable \(k\), we can factor it out of the sum:

\(S = n \sum_{k=1}^n \binom{n-1}{k-1}\)

Let's change the index of summation. Let \(j = k-1\). When \(k=1\), \(j=1-1=0\). When \(k=n\), \(j=n-1\). The sum becomes:

\(S = n \sum_{j=0}^{n-1} \binom{n-1}{j}\)

Recall the Binomial Theorem sum formula: \(\sum_{j=0}^m \binom{m}{j} = 2^m\). In our sum, we have \(m = n-1\). So, the sum \(\sum_{j=0}^{n-1} \binom{n-1}{j}\) is equal to \(2^{n-1}\).

Substitute this back into the expression for \(S\):

\(S = n \cdot 2^{n-1}\)

Method 2: Using Differentiation of Binomial Series Expansion

Consider the given binomial expansion:

\((1+x)^n = 1 + T_1x + T_2x^2 + T_3x^3 + \dots + T_nx^n\)

As established, \(T_k = \binom{n}{k}\) for \(k \ge 1\), and the constant term is \(\binom{n}{0} = 1\). The expansion can be written as:

\((1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k\)

To obtain terms like \(k \cdot T_k\) (or \(k \cdot \binom{n}{k}\)), we can differentiate the series with respect to \(x\). Differentiating \(T_k x^k\) with respect to \(x\) gives \(k T_k x^{k-1}\).

Differentiate both sides of the equation \((1+x)^n = 1 + T_1x + T_2x^2 + T_3x^3 + \dots + T_nx^n\) with respect to \(x\):

Left side: \(\frac{d}{dx} (1+x)^n = n(1+x)^{n-1}\)

Right side: \(\frac{d}{dx} (1 + T_1x + T_2x^2 + T_3x^3 + \dots + T_nx^n)\)

This is the sum of the derivatives of each term:

\(= \frac{d}{dx}(1) + \frac{d}{dx}(T_1x) + \frac{d}{dx}(T_2x^2) + \frac{d}{dx}(T_3x^3) + \dots + \frac{d}{dx}(T_nx^n)\)

\(= 0 + T_1(1) + T_2(2x) + T_3(3x^2) + \dots + T_n(nx^{n-1})\)

\(= T_1 + 2T_2x + 3T_3x^2 + \dots + nT_nx^{n-1}\)

Equating the results from differentiating both sides:

\(n(1+x)^{n-1} = T_1 + 2T_2x + 3T_3x^2 + \dots + nT_nx^{n-1}\)

The sum we are looking for is \(T_1 + 2T_2 + 3T_3 + \dots + nT_n\). This is the expression on the right side when we set \(x=1\).

Substitute \(x=1\) into the equation:

\(n(1+1)^{n-1} = T_1 + 2T_2(1) + 3T_3(1)^2 + \dots + nT_n(1)^{n-1}\)

\(n(2)^{n-1} = T_1 + 2T_2 + 3T_3 + \dots + nT_n\)

Thus, the sum \(T_1 + 2T_2 + 3T_3 + \dots + nT_n\) is equal to \(n2^{n-1}\).

Summary of Binomial Coefficient Sum Result

Both methods, using combinatorial identity and using differentiation, yield the same result for the required sum involving binomial expansion coefficients.

Revision Table: Key Binomial Concepts Review

Additional Information: Further Sums and Properties

Studying sums involving binomial coefficients often involves applying the Binomial Theorem itself, using identities, or employing calculus techniques like differentiation or integration on series expansions. Here are a few more examples of related sums and properties:

• Sum of alternate binomial coefficients: \(\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} = \sum_{k=0}^n (-1)^k \binom{n}{k} = (1-1)^n = 0\) for \(n \ge 1\).
• Sum weighted by powers of a number: \(\sum_{k=0}^n \binom{n}{k} y^k = (1+y)^n\). For example, \(\sum_{k=0}^n \binom{n}{k} 2^k = (1+2)^n = 3^n\).
• Sum of squares of binomial coefficients: \(\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}\). This arises from considering the coefficient of \(x^n\) in the expansion of \((1+x)^n (x+1)^n = (1+x)^{2n}\).

Problems involving sums of coefficients often require recognizing which form of the binomial expansion or its derivatives/integrals, evaluated at a specific point (like \(x=1\) or \(x=-1\)), will produce the desired sum.