Key Concepts and Formulas

This problem utilizes two fundamental properties of binomial coefficients:

1. Symmetry Property: For any non-negative integers $n$ and $r$ such that $0 \le r \le n$, the binomial coefficient $^nC_r$ is equal to $^nC_{n-r}$. $^nC_r = ^nC_{n-r}$ Why this is important: This property implies that the coefficients in a binomial expansion are symmetric. For example, in $(x+y)^{25}$, the coefficient of $x^0y^{25}$ ($^{25}C_0$) is the same as the coefficient of $x^{25}y^0$ ($^{25}C_{25}$). This symmetry is key to simplifying sums involving binomial coefficients.

2. Sum of Binomial Coefficients: The sum of all binomial coefficients for a given $n$ is $2^n$. \sum_{r=0}^{n} ^nC_r = ^nC_0 + ^nC_1 + \dots + ^nC_n = 2^n Why this is important: This identity directly comes from setting $x=1$ in the binomial expansion of (1+x)^n = \sum_{r=0}^{n} ^nC_r x^r. It provides a simple closed form for sums of this type.

The core strategy to solve this problem involves a clever application of the symmetry property: reversing the series and adding it to the original series. This technique is particularly useful when the binomial coefficients are multiplied by terms that form an Arithmetic Progression (A.P.).

Step-by-Step Solution

Step 1: Understand the Given Series Let the given sum be $S$. $S = 1 \cdot ^{25}C_0 + 5 \cdot ^{25}C_1 + 9 \cdot ^{25}C_2 + \dots + (101) \cdot ^{25}C_{25}$ Explanation: We begin by carefully examining the structure of the series. The terms involve binomial coefficients of the form $^{25}C_r$, where $r$ ranges from $0$ to $25$. Each binomial coefficient is multiplied by a numerical coefficient: $1, 5, 9, \dots, 101$.

Why this step is taken: Understanding the pattern is crucial for choosing the right method. We observe that the multipliers $1, 5, 9, \dots, 101$ form an Arithmetic Progression (A.P.). Let's find the general term of this A.P. for the $r$-th term, assuming $r$ starts from $0$: The first term (for $r=0$) is $a_0 = 1$. The common difference $d = a_1 - a_0 = 5 - 1 = 4$. So, the general term of the A.P. for $r=0, 1, 2, \dots, 25$ is $a_r = a_0 + r \cdot d = 1 + 4r$. Let's verify the last term: for $r=25$, $a_{25} = 1 + 4(25) = 1 + 100 = 101$. This matches the last coefficient in the series. Thus, the series can be written in summation notation as: $S = \sum_{r=0}^{25} (4r+1) \cdot ^{25}C_r$

Step 2: Rewrite the Series using the Symmetry Property Now, we apply the symmetry property of binomial coefficients: $^nC_r = ^nC_{n-r}$. For $n=25$, this means: $^{25}C_0 = ^{25}C_{25}$ $^{25}C_1 = ^{25}C_{24}$ ... $^{25}C_{25} = ^{25}C_0$

We can rewrite the sum $S$ by expressing it in terms of $r$ values going from $25$ down to $0$, and using the corresponding coefficients from the A.P. in reverse order. Alternatively, and more directly for this method, we can write out the series in the usual $r=0$ to $r=25$ order, but pair each $^{25}C_r$ with the A.P. coefficient that would have been with $^{25}C_{25-r}$ in the original series.

Let's write the original series (1): (1) $S = 1 \cdot ^{25}C_0 + 5 \cdot ^{25}C_1 + \dots + 97 \cdot ^{25}C_{24} + 101 \cdot ^{25}C_{25}$

Now, let's write $S$ again, but for each term $^{25}C_r$, we associate it with the coefficient $a_{25-r}$ from the A.P. (i.e., the coefficient that was originally with $^{25}C_{25-r}$). The A.P. coefficient for index $25-r$ is $1 + 4(25-r) = 1 + 100 - 4r = 101 - 4r$. So, the series can be expressed as: $S = \sum_{r=0}^{25} (101 - 4r) \cdot ^{25}C_r$ Writing this out term by term: (2) $S = (101-4(0)) \cdot ^{25}C_0 + (101-4(1)) \cdot ^{25}C_1 + \dots + (101-4(25)) \cdot ^{25}C_{25}$ (2) $S = 101 \cdot ^{25}C_0 + 97 \cdot ^{25}C_1 + \dots + 5 \cdot ^{25}C_{24} + 1 \cdot ^{25}C_{25}$

Explanation: By applying the symmetry $^nC_r = ^nC_{n-r}$, we effectively reverse the sequence of the binomial coefficients. To maintain equality, the numerical coefficients must also be taken in reverse order. The term with $^{25}C_0$ in the original series had coefficient $1$. When we rewrite, $^{25}C_0$ is equivalent to $^{25}C_{25}$, which had coefficient $101$. So, the new series assigns the original last coefficient to the first binomial term, the original second-to-last to the second, and so on.

Step 3: Add the Two Expressions for $S$ Now, we add equation (1) and equation (2) term by term: $2S = (1 \cdot ^{25}C_0 + 5 \cdot ^{25}C_1 + \dots + 101 \cdot ^{25}C_{25}) + (101 \cdot ^{25}C_0 + 97 \cdot ^{25}C_1 + \dots + 1 \cdot ^{25}C_{25})$ Combining corresponding terms: $2S = (1+101) \cdot ^{25}C_0 + (5+97) \cdot ^{25}C_1 + \dots + (101+1) \cdot ^{25}C_{25}$ Explanation: This is the core trick of the method. By adding the series to its "reversed" form, the arithmetic progression components simplify significantly. Let's check the sum of the numerical coefficients for each term: For $r=0$: $1 + 101 = 102$ For $r=1$: $5 + 97 = 102$ For $r=2$: $9 + 93 = 102$ ... For $r=25$: $101 + 1 = 102$

Indeed, the sum of the numerical coefficients for each pair of terms is constant and equal to $102$. So, we can factor out $102$: $2S = 102 \cdot (^{25}C_0 + ^{25}C_1 + ^{25}C_2 + \dots + ^{25}C_{25})$ 2S = 102 \sum_{r=0}^{25} ^{25}C_r

Step 4: Apply the Sum of Binomial Coefficients Identity We use the identity \sum_{r=0}^{n} ^nC_r = 2^n. For our case, $n=25$, so: \sum_{r=0}^{25} ^{25}C_r = 2^{25} Explanation: This identity allows us to simplify the sum of all binomial coefficients into a simple power of 2. Substitute this back into the equation for $2S$: $2S = 102 \cdot 2^{25}$ Now, divide both sides by 2 to find $S$: $S = \frac{102 \cdot 2^{25}}{2}$ $S = 51 \cdot 2^{25}$

Step 5: Determine the Value of k The problem states that $S = 2^{25} \cdot k$. We have calculated $S = 51 \cdot 2^{25}$. By comparing these two expressions for $S$: $51 \cdot 2^{25} = k \cdot 2^{25}$ Explanation: We equate our calculated value of $S$ with the form given in the question to solve for the unknown $k$. Dividing both sides by $2^{25}$ (since $2^{25} \neq 0$): $k = 51$

Tips and Common Mistakes to Avoid

• Identifying the A.P.: Always verify that the coefficients multiplying the binomial terms indeed form an Arithmetic Progression. This method fundamentally relies on that structure.
• Correctly Reversing the Series: When creating the second series using $^nC_r = ^nC_{n-r}$, ensure that the numerical coefficients are correctly paired. The coefficient for $^{n}C_r$ in the original series is $a_r$. In the reversed series, the coefficient for $^{n}C_r$ should be $a_{n-r}$.
• Don't Forget the $2S$: Remember that by adding two series (the original and the reversed), you obtain $2S$, so the final step will always involve dividing the sum by 2.
• General Term of A.P.: Be careful when deriving the general term for the A.P. Consistency in indexing (whether $r$ starts from $0$ or $1$) is crucial.

Summary and Key Takeaway

This problem demonstrates an elegant and efficient technique for evaluating sums of the form $\sum_{r=0}^{n} (Ar+B) \cdot ^nC_r$, where the numerical coefficients form an Arithmetic Progression. By exploiting the symmetry property of binomial coefficients ($^nC_r = ^nC_{n-r}$) and rewriting the sum in reverse, we can pair terms such that the arithmetic progression components sum to a constant. This allows us to factor out that constant and reduce the problem to the straightforward sum of all binomial coefficients, which is $2^n$. This method is a powerful tool in solving various problems involving binomial series coupled with arithmetic progressions.