Q67.The number of terms in the expansion of (1 + x)101(1 โx + x2) 100 in powers of x is (1) 301 (2) 302 (3) 101 (4) 202
What This Question Tests
This problem tests the ability to simplify algebraic expressions involving binomial expansions by using common factorization formulas and then determine the number of distinct terms in the resulting polynomial.
Concepts Tested
Formulas Used
(a+b)(a^2-ab+b^2) = a^3+b^3
Binomial expansion of (1+x)^n
๐ NCERT Sections This Tests
1.3 โ Define The Following Terms:
Chemistry Class 11 ยท Chapter 1
1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
1.1 โ Define The Term Solution. How Many Types Of Solutions Are Formed? Write Briefly
Chemistry Class 11 ยท Chapter 1
1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
5.12 โ Write All The Geometrical Isomers Of [Pt(Nh3)(Br)(Cl)(Py)] And How Many Of
Chemistry Class 11 ยท Chapter 5
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
๐ Question Details
- Chapter
- Binomial Theorem
- Topic
- Binomial expansion
- Year
- 2014
- Shift
- 09 Apr Online
- Q Number
- Q67
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 8: Binomial Theorem
More from this Chapter
Q89.In the binomial expansion of (a โb)n, n โฅ5 , the sum of 5th and 6th terms is zero, then ab equals (1) 5 (2) 6 nโ4 nโ5 (3) nโ5 (4) nโ4 6 5
Q90.The sum of the series 20C0 โ20C1 + 20C2 โ20C3 + โฆ โโฆ + 20C10 is (1) โ20C10 (2) 12 20C10 (3) 0 (4) 20C10
Q77.Statement-1: โnr=0(r + 1)nCr = (n + 2)2nโ1 Statement -2: โnr=0(r + 1)nCrxr = (1 + x)n + nx(1 + x)nโ1 . JEE Main 2008 JEE Main Previous Year Paper (1) Statement โ1 is false, Statement โ2 is true (2) Statement โ1 is true, Statement โ2 is true, Statement โ2 is a correct explanation for Statement โ1 (3) Statement โ1 is true, Statement โ2 is true; (4) Statement โ1 is true, Statement โ2 is false. Statement โ2 is not a correct explanation for Statement โ1.
Q65.The remainder left out when 82n โ(62)2n+1 is divided by 9 is (1) 0 (2) 2 (3) 7 (4) 8