
Introduce the binomial theorem and the first principle of mathematical induction, defining mathematical statements and distinguishing true and false examples, and explore sum formulas for n, squares, and cubes.
Learn the first principle of mathematical induction: establish a base case, assume the statement for k, and prove the induction step. Apply to the sum 1+2+...+n = n(n+1)/2.
Explore binomial coefficients in the binomial expansion, showing that the sum of coefficients equals 2^n and that the sums of even and odd coefficients are equal.
Learn to find coefficients in binomial expansions using the general term T_{r+1}=nCr a^{n-r} b^r. Determine the coefficient of x in (x+1)^3 and apply r to get the desired term.
Prove the binomial theorem for positive integral index using mathematical induction, establishing the base case and the inductive step with binomial coefficients.
Learn how to reduce the binomial's first term to unity by using the one plus x form of the binomial theorem, with practical example calculations.
Explore the binomial theorem and binomial expressions, from simple cases to higher powers, using Pascal's triangle to identify coefficients and connections to combinations.
Derive the general term to expand powers of binomials using binomial coefficients, showing the pattern of terms C(n,r) a^(n-r) b^r and verifying with (x+y)^n.
Learn to find the middle term(s) in binomial expansions using the binomial theorem. Even n gives a single middle term, while odd n gives two, with examples.
Discover a new form of the binomial theorem for expansion, pattern, and application using one plus x to the n, with comparisons to the original form.
Explore the binomial theorem in a new form by expanding (1+x)^(1/2) and (1-x)^(1/2), revealing fractional-power series and coefficients like 1/2 and -1/8.
Introduce the generalized binomial expansion for non-positive exponents, deriving (1+x)^-1 as 1 - x + x^2 - x^3 + …, and apply it to compute reciprocals like 1/1.1.
Applying the binomial theorem to approximate sqrt(99) by expressing it as (100 − 1)^(1/2) and expanding around 100, achieving four-decimal accuracy.
Demonstrate a mathematical induction proof that 3^n minus 1 is divisible by seven for all natural numbers, including base case, inductive assumption, and the induction step.
expand the binomial (2x^2+3)^4 using the binomial theorem, derive terms for k=0 to 4, and obtain the expanded form 16x^8 + 96x^6 + 216x^4 + 216x^2 + 81.
Use binomial theorem to expand (1.1)^5 by computing binomial coefficients (5C0 to 5C5) and summing terms to obtain 1.61051.
Use the binomial theorem to expand ten minus point one cubed, collect coefficients, and obtain 970.29299 for 9.9 cubed.
Apply the binomial theorem to (x^2 - 4/x)^{11} to find the fifth term. Use T_{r+1}=C(11,r)a^{11-r}b^{r} with a=x^2 and b=-4/x to get 84480 x^{10}.
Compute the coefficient of x^9 in the expansion of (1/x + x^2)^18 using the binomial theorem; set -18+3r=9 to obtain r=9, giving 18C9=48620.
Identify the term independent of x in the binomial expansion with n=9, using binomial coefficients and r, including 9C3, to isolate the constant term.
Derive the first four terms of the given binomial expansion using the binomial theorem. Demonstrate the coefficients and powers, including 1, x/3, 2/9 x^2, and 14/27 x^3.
Demonstrate summing binomial coefficients c0 through c9 using the binomial theorem and verify the sum equals 2^9, highlighting the double formula.
Explore level-2 numericals on mathematical induction and binomial theorem through solved questions, covering induction proofs, the binomial middle terms, and practical expansion techniques.
Explore level-2 numericals on mathematical induction and the binomial theorem, including expansion comparisons, sum of coefficients, and coefficient extraction.
Principle of Mathematical Induction
Process of the proof by induction −
Motivating the application of the method by looking at natural numbers as the least inductive subset of real numbers
The principle of mathematical induction and simple applications
Binomial Theorem
History
Statement and proof of the binomial theorem for positive integral indices
Pascal's triangle
General and middle term in binomial expansion
Simple applications
SUMMARY
Principle of Mathematical Induction
1. One key basis for mathematical thinking is deductive reasoning. In contrast to deduction, inductive reasoning depends on working with different cases and developing a conjecture by observing incidences till we have observed each and every case. Thus, in simple language we can say the word ‘induction’ means the generalisation from particular cases or facts.
2. The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as P(n) associated with positive integer n, for which the correctness for the case n = 1 is examined. Then assuming the truth of P(k) for some positive integer k, the truth of P (k+1) is established.
3. Property (i) - is simply a statement of fact. There may be situations when a statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall verify the result for n = 4, i.e., P(4).
4. Property (ii) - is a conditional property. It does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1.
Binomial Theorem
1. A triangle with 1 at the top vertex and running down the two slanting sides. This array of numbers is known as Pascal’s triangle, after the name of French mathematician Blaise Pascal. It is also known as Meru Prastara by Pingla.
2. The coefficients nCr occuring in the binomial theorem are known as binomial coefficients.
3. There are (n+1) terms in the expansion of (a+b) n, i.e., one more than the index.
4. In the successive terms of the expansion the index of a goes on decreasing by unity. It is n in the first term, (n–1) in the second term, and so on ending with zero in the last term. At the same time the index of b increases by unity, starting with zero in the first term, 1 in the second and so on ending with n in the last term.
5. In the expansion of (a+b) raise to n , the sum of the indices of a and b is n + 0 = n in the first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion.