Binomial Theorem — Flashcards
ICSE · Class 11 · Mathematics
29 flashcards for Binomial Theorem (ICSE Class 11 Mathematics) to test yourself on key terms and facts. Sample: "Find the 4th term in the expansion of"
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Find the 4th term in the expansion of (x + 2)^5.
Answer
Use the general term: T(r+1) = nCr x^(n-r) y^r. Here, n = 5, x = x, y = 2. For the 4th term, r + 1 = 4 so r = 3. T4 = 5C3 x^(5-3) 2^3 = 10 x^2 × 8 = 80x^2. Answer: 80x^2…
Find the 2nd term in the expansion of (2x - 3)^6.
Answer
Use T(r+1) = nCr a^(n-r) b^r with (2x - 3)^6 = [2x + (-3)]^6. For the 2nd term, r = 1. T2 = 6C1 (2x)^5 (-3)^1 = 6 × 32x^5 × (-3) = -576x^5. Answer: -576x^5…
Find the coefficient of x^3 in the expansion of (x - x^2)^10.
Answer
Write (x - x^2)^10 = [x + (-x^2)]^10. General term: T(r+1) = 10Cr x^(10-r) (-x^2)^r = 10Cr (-1)^r x^(10-r+2r) = 10Cr (-1)^r x^(10+r). For x^3, we need 10 + r = 3, which gives r = -7. That is not possi…
Find the term independent of x in (3x - 2/x^2)^15.
Answer
Write (3x - 2/x^2)^15 = [3x + (-2/x^2)]^15. General term: T(r+1) = 15Cr (3x)^(15-r) (-2/x^2)^r = 15Cr 3^(15-r) (-2)^r x^(15-r-2r) = 15Cr 3^(15-r) (-2)^r x^(15-3r). For the term independent of x, power…
What is the general term of (x + y)^n?
Answer
The (r + 1)th term is: T(r+1) = nCr x^(n-r) y^r. Meaning of symbols: - n = positive integer exponent - r = term index starting from 0 to n - x^(n-r) gives the power of x - y^r gives the power of y Qui…
Why does the number of terms in (x + y)^n always equal n + 1?
Answer
In the expansion of (x + y)^n, the term index r runs from 0 to n. That gives: T1, T2, ..., T(n+1) So the total number of terms is n + 1. Example: For (x + y)^4, terms are: x^4, 4x^3y, 6x^2y^2, 4xy^3, …
Find the 5th term from the beginning in (x + a)^n using the general term formula.
Answer
The general term is T(r+1) = nCr x^(n-r) a^r. For the 5th term, r + 1 = 5, so r = 4. Hence, T5 = nC4 x^(n-4) a^4. This is the 5th term from the beginning, not from the end.
Find the 4th term from the end of (x + a)^n.
Answer
The rth term from the end is T(n-r+2). For r = 4, 4th term from the end = T(n-4+2) = T(n-2). Now use the general term: T(n-2) = nC(n-3) x^3 a^(n-3). Using symmetry of combinations, T(n-2) = nC3 a^(n-3…
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