Understanding Binomial Expansion and Coefficients

The question deals with the binomial expansion of \((1 + x)^n\). The standard binomial theorem states that for any positive integer \(n\), the expansion of \((a+b)^n\) is given by:

\((a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n} a^0 b^n\)

For the expression \((1+x)^n\), we substitute \(a=1\) and \(b=x\) into the binomial theorem formula:

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

Since \(1^k = 1\) for any power \(k\) and \(x^0 = 1\), this simplifies to:

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

The question gives the expansion as \((1 + x)^n = 1 + T_1x + T_2x^2 + T_3x^3 + \dots + T_nx^n\). By comparing the coefficients of corresponding powers of \(x\) in both expansions, we can identify the terms \(T_k\).

• The constant term (coefficient of \(x^0\)) is \(\binom{n}{0}\) which is equal to 1. This matches the given expansion.
• The coefficient of \(x\) is \(T_1\) in the given expansion and \(\binom{n}{1}\) from the binomial theorem. So, \(T_1 = \binom{n}{1}\).
• The coefficient of \(x^2\) is \(T_2\) in the given expansion and \(\binom{n}{2}\) from the binomial theorem. So, \(T_2 = \binom{n}{2}\).
• The coefficient of \(x^3\) is \(T_3\) in the given expansion and \(\binom{n}{3}\) from the binomial theorem. So, \(T_3 = \binom{n}{3}\).
• ... and so on, up to the coefficient of \(x^n\), which is \(T_n\) in the given expansion and \(\binom{n}{n}\) from the binomial theorem. So, \(T_n = \binom{n}{n}\).

Thus, we have \(T_k = \binom{n}{k}\) for \(k = 1, 2, \dots, n\). The question asks for the sum \(T_1 + T_2 + T_3 + \dots + T_n\). This sum is \(\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \dots + \binom{n}{n}\).

Sum of Binomial Coefficients Calculation

We know a very useful property of binomial coefficients: the sum of all coefficients in the expansion of \((1+x)^n\) is obtained by setting \(x=1\). If we substitute \(x=1\) into the expansion \((1+x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \dots + \binom{n}{n} x^n\), we get:

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

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

This shows that the sum of all binomial coefficients from \(k=0\) to \(n\) for a given \(n\) is \(2^n\). We want to find the sum \(T_1 + T_2 + \dots + T_n\), which corresponds to \(\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}\).

The sum of all coefficients is \(\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n\).

We can write this as:

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

Recognizing that \(\binom{n}{0} = 1\), and the part in the parenthesis is the sum \(T_1 + T_2 + \dots + T_n\), we have:

\(1 + (T_1 + T_2 + \dots + T_n) = 2^n\)

To find the required sum, we can subtract 1 from both sides of the equation:

\(T_1 + T_2 + \dots + T_n = 2^n - 1\)

Conclusion: Value of T1 + T2 + ... + Tn

Based on the binomial expansion and the property of the sum of binomial coefficients, the sum \(T_1 + T_2 + \dots + T_n\) is equal to \(2^n - 1\). This is because the total sum of coefficients in \((1+x)^n\) is \(2^n\), and this total sum includes the first term (the constant term, coefficient of \(x^0\)) which is \(T_0 = \binom{n}{0} = 1\). The sum \(T_1 + \dots + T_n\) is the sum of all coefficients except the first one.

Revision Table: Key Concepts Summary

Additional Information: Properties of Binomial Coefficients

Binomial coefficients, denoted by \(\binom{n}{k}\) or \(^nC_k\), represent the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to the order of selection. They appear in the binomial expansion and have several interesting properties:

• Symmetry Property: \(\binom{n}{k} = \binom{n}{n-k}\). This means the coefficient of \(x^k\) is the same as the coefficient of \(x^{n-k}\) in the expansion of \((1+x)^n\). For example, \(T_1 = \binom{n}{1}\) and \(T_{n-1} = \binom{n}{n-1}\), and \(\binom{n}{1} = \binom{n}{n-1}\).
• Pascal's Identity: \(\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\) for \(1 \le k \le n-1\). This identity forms the basis of Pascal's triangle, where each number is the sum of the two numbers directly above it.
• Sum of Alternating Coefficients: \(\sum_{k=0}^n (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} = 0\) for \(n \ge 1\). This is obtained by setting \(x=-1\) in the expansion of \((1+x)^n\).
• Sum of even/odd coefficients: From the total sum (\(2^n\)) and the alternating sum (0), we can find the sum of even-indexed coefficients and odd-indexed coefficients. The sum of even-indexed coefficients (\(\binom{n}{0} + \binom{n}{2} + \dots\)) equals the sum of odd-indexed coefficients (\(\binom{n}{1} + \binom{n}{3} + \dots\)), and both are equal to \(2^{n-1}\) for \(n \ge 1\).

Understanding these properties is helpful when working with binomial expansions and coefficient sums.