Airforce X Y / Indian Navy SSR Exam > Airforce X Y / Indian Navy SSR Videos > Mathematics for Airmen Group X > Examples (NCERT) : Part 7 - Binomial Theorem
Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR
|
149 videos|192 docs|197 tests
|
FAQs on Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture - Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR
| 1. What is the binomial theorem? |  |
Ans. The binomial theorem is a mathematical formula that allows us to expand the powers of a binomial expression. It states that for any positive integer n, the n-th power of a binomial expression (a + b) can be expanded as the sum of n+1 terms. The terms in the expansion are obtained by raising the first term (a) to decreasing powers and the second term (b) to increasing powers, with the coefficients given by the binomial coefficients.
| 2. How do you find the coefficient of a specific term in the expansion of a binomial expression? | |
Ans. To find the coefficient of a specific term in the expansion of a binomial expression, we use the formula for binomial coefficients. The coefficient of the term with the power of a raised to p and b raised to q is given by the binomial coefficient C(n, k), where n is the power of the binomial expression, and k is the power of a in the specific term. The binomial coefficient C(n, k) is calculated using the formula C(n, k) = n! / (k! * (n-k)!), where ! denotes factorial.
| 3. What is the significance of the binomial theorem in combinatorics? | |
Ans. The binomial theorem plays a significant role in combinatorics, which is the branch of mathematics concerned with counting and arranging objects. The binomial coefficients obtained from the binomial theorem represent the number of ways to choose a certain number of objects from a larger set without regard to their order. These coefficients help in solving various combinatorial problems, such as counting the number of combinations, permutations, and arrangements of objects.
| 4. Can the binomial theorem be used to find the value of irrational or non-integer exponents? | |
Ans. No, the binomial theorem is applicable only for positive integer exponents. The binomial theorem is derived based on the pattern observed in the expansion of binomial expressions with positive integer exponents. It does not hold true for irrational or non-integer exponents. For such cases, other mathematical techniques like calculus or logarithms are used to find the value of the expression.
| 5. How can the binomial theorem be applied in probability theory? | |
Ans. The binomial theorem finds applications in probability theory, particularly in the calculation of probabilities in binomial distributions. The binomial distribution represents the probability of obtaining a certain number of successes in a fixed number of independent Bernoulli trials, where each trial has two possible outcomes (success or failure). The binomial coefficients from the binomial theorem are used to calculate the probabilities associated with different outcomes in binomial experiments.
Text Transcript from Video
hello friends this video on my name is
theorem part 7 is brought to you by exam
fear calm no more fear from exam before
watching this video you should have
washed my name a theorem part 1 to part
6 let's take one example that is little
difficult so we have to expand this x
squared plus 4x the power sector will do
this so this we can say just nothing but
sigma art from 0 to 7 because n 0 7 7
see our current and use it X to the
power 4 X square to the power n x
squared the power R into we have 4 by X
4 over X to the power n minus 1 you can
also say X square to the power n minus 4
into voiced form X bar on but let's take
this R Bar n minus 1 sum is also + R
plus n minus R value in this is the
expansion of this so now let's do the
actual expansion you can say I is going
to 0 so we get 7 c 0 u X square to the
power 0 because R is produced 0 x
squared the power 0 into 4 by X to the
power n minus 0 that is n is here 7 self
it's thinner 7 + 7 c 1 x squared will
increase the value 0 to 1 and we will
decrease the value of 7 by 1 this
becomes 6 or so 7 minus R is good to 1 3
come 6 plus now or is equal to 2 7 C T X
square to the power 2 because R is 2
plus 4 by 7 4 by X 7 minus 2 is 35
plus 7c 3x squared the base 3 into 1 by
X to the power 4 because 7 minus 3 is
equal to 4 plus 7c 4 into X square to
the power 4 into 4 by X to the bar
because this is 4 7 minus 4 is equal to
3 plus 7c 5x squared to the power 5 X
where the power are our 25 into 4 by X
to the power 7 minus 5 that is 2 7 c 6x
squared to the bars 6 into 4 by X to the
power 7-6 31 plus 7c 7x squared to the
power 7 this is the total expansion of
this will now wait this so what we get
is this is a 1 2 7 C 1 7 C naught is
equal to 1 x squared to the power 0 is
equal to 1 so we get 4 by X to the power
7 plus 7 C 1 is going to 7 into X square
into 4 by 6 X to the power 6 plus 7c t 7
C 2 is going to 21 you solve this you
get 21 into X to the power 4 into 4 to
the power 5 by X to the power 5 plus 7 C
3 is good but if I'm reading 5 into X to
the power 6 into 4 to the power 4 by X
to the power 4 please note that 76135 we
have learned is a message up to 74 also
becomes 35 so we say 25 into X to the
power 7 C 4
eight into work will bar 3 by X cube
plus 75 this term will take 7c v v 121 X
to the power n x squared plus 5 into 4
to the power by X to the power plus 7
into X square to the power 6 that
becomes X got 12 into 4 1 into H to the
power 8 plus this becomes 7 into 4 into
X to the power 11 plus X to the power 14
so if you see the power is increasing by
3 H to the power minus 7 because if you
move it up X to the power minus 1 X to
the power minus 1 this become X to the
power 2 2 plus 3 is 5 X to power 5 this
becomes equal 5 + 3 8 this becomes 11
this becomes 40 so this is now answer
for X square plus 4x by 7 if you expand
this so you get this as the easy so we
have an example where we have to think a
little bit we are asked to find 98 to
the power 5 what we can do we have been
told that we have we know the formula of
a plus B to the power n or a minus B
power n so we know all this formula but
we don't know now I need to power 5 so
we can also say 98 is equal to hundred
minus 2 and 98 for 5 is going 200 minus
2 but fine and now using one binomial
theorem we can solve this so let's see
98 the power 5 is equal to 100 minus 2
is gone fine to expand this we'll get
this summation r is equal to 0 to 5 5 C
our build a hundred first hundred to the
power R into minus 2 to the power 5
minus R this is the formula we get for
NCR expansion what so now let's solve
this
so this become our is good 2 0 5 c 0 100
to the power 0 is equal to again 0 into
minus 2 power 5 plus 5 C 1 into hundred
to the power 1 and minus 2 to the power
4 plus 5 C 2 5c to 100th part 2 into
minus 2 to the power 3 plus 5 C 300 to
the power 3 into minus 2 to the power
this guy will decrease 2 plus 5 C 4 into
100 to the power 4 into minus 2 to the
power 1 plus 5 C 5 into 100 to the power
5 into minus to the power 0 this is the
expansion so expand this you get five
C's respond to one in the power 5 and to
the power Phi is equal to minus 32x
there so we'll say minus 32 and then Phi
C 1 is equal to 5 into 100 into minus 15
power 4 is equal to 16 plus 5 c 2 5
scene is good let's find out this year
we have done combination Phi C 2 is
equal to 5 permutation by 3 permutation
addition this bill to 5 into 4 into 3
formulation by 3 permutation into and
this is gone this is T equal to 10 so 5
C 2 is equal to 10 so I said n into 100
square that is 100 into hundred into
minus 2 whole cube is going to minus 8
this becomes a minus a minus a plus 5 C
3 also becomes same that is 10 into this
is 100 Q into minus 2 whole square is
equal to 4 plus 5 C 4 is equal to 1 into
102 the bar 4 2 2 3 4 in it 2 to the
power 1 into 2 minus 1 so we'll say
minus here and plus we see Phi C Phi is
equal to 1 10 to power 5 is equal to 3 4
5 times 2 0 into minus 2 to the power 0
is equal to 1 so you got this so if you
solve this what you get is 9:03 9:07 -
is the answer what we have done so we
have converted 90 to the power 5 into
100 - 2 / 5 we have applied the binomial
theorem and expanded this to get the
answer let's take one more example where
we think little more so we are asked to
tell which one is greater this artists
finding this power is difficult so we
can use the power of my limit we can
also write one point zero one is equal
to one plus point zero 1 to the power 1
2 3 1 by 16 so now I can expand this
expand this what we'll see is this guy
is nothing but 1 1 2 3 4 5 6 0 C naught
into 1 to the power 1 2 3 4 5 6 into
point 0 1 to the power 0 Plus this time
6 times this C 1 into 1 to the power 9
the subtract this 1 2 3 4 5 6 3 3 3 6 7
3 what is extra so you want this to get
smoother this is not there in 2.0 1/3
power 1 plus dollar when you solve this
100 C 0 is equal to 1 into 1 to the
power this cut the 1 into point zero 1
plus this C 1 this itself becomes this
much 1 2
or 6 into 1 to the power this is good to
one in 2.0 1 to the power 1 is going to
on by Amtrak plus again some constant
will say constant this also will receive
as a constant so this by this cancel so
this becomes 10,000 plus some constant
so that means this is always greater
than 10,000 so when we are supposed to
compare this and this so obviously this
is little because we have proved that 1
point 0 1 to the power 6 time this this
is nothing but n thousand plus some
constant so when he asked this on this
tenth of the mission is greater
obviously this greater because this is
equal to this plus embossing so then say
that this one is 3 so what we have done
we have written 1 point 0 1 to the power
n in the form of 1 plus 1 point 0 to the
power n we expanded this and we found
that this is nothing but 10,000 plus
constant that if this number is greater
than n thousand thank you visit exam
care.com
to watch free educational videos try
free online test if the best quality
study material study from the best tutor
and mentors and much forum times once
again
theorem part 7 is brought to you by exam
fear calm no more fear from exam before
watching this video you should have
washed my name a theorem part 1 to part
6 let's take one example that is little
difficult so we have to expand this x
squared plus 4x the power sector will do
this so this we can say just nothing but
sigma art from 0 to 7 because n 0 7 7
see our current and use it X to the
power 4 X square to the power n x
squared the power R into we have 4 by X
4 over X to the power n minus 1 you can
also say X square to the power n minus 4
into voiced form X bar on but let's take
this R Bar n minus 1 sum is also + R
plus n minus R value in this is the
expansion of this so now let's do the
actual expansion you can say I is going
to 0 so we get 7 c 0 u X square to the
power 0 because R is produced 0 x
squared the power 0 into 4 by X to the
power n minus 0 that is n is here 7 self
it's thinner 7 + 7 c 1 x squared will
increase the value 0 to 1 and we will
decrease the value of 7 by 1 this
becomes 6 or so 7 minus R is good to 1 3
come 6 plus now or is equal to 2 7 C T X
square to the power 2 because R is 2
plus 4 by 7 4 by X 7 minus 2 is 35
plus 7c 3x squared the base 3 into 1 by
X to the power 4 because 7 minus 3 is
equal to 4 plus 7c 4 into X square to
the power 4 into 4 by X to the bar
because this is 4 7 minus 4 is equal to
3 plus 7c 5x squared to the power 5 X
where the power are our 25 into 4 by X
to the power 7 minus 5 that is 2 7 c 6x
squared to the bars 6 into 4 by X to the
power 7-6 31 plus 7c 7x squared to the
power 7 this is the total expansion of
this will now wait this so what we get
is this is a 1 2 7 C 1 7 C naught is
equal to 1 x squared to the power 0 is
equal to 1 so we get 4 by X to the power
7 plus 7 C 1 is going to 7 into X square
into 4 by 6 X to the power 6 plus 7c t 7
C 2 is going to 21 you solve this you
get 21 into X to the power 4 into 4 to
the power 5 by X to the power 5 plus 7 C
3 is good but if I'm reading 5 into X to
the power 6 into 4 to the power 4 by X
to the power 4 please note that 76135 we
have learned is a message up to 74 also
becomes 35 so we say 25 into X to the
power 7 C 4
eight into work will bar 3 by X cube
plus 75 this term will take 7c v v 121 X
to the power n x squared plus 5 into 4
to the power by X to the power plus 7
into X square to the power 6 that
becomes X got 12 into 4 1 into H to the
power 8 plus this becomes 7 into 4 into
X to the power 11 plus X to the power 14
so if you see the power is increasing by
3 H to the power minus 7 because if you
move it up X to the power minus 1 X to
the power minus 1 this become X to the
power 2 2 plus 3 is 5 X to power 5 this
becomes equal 5 + 3 8 this becomes 11
this becomes 40 so this is now answer
for X square plus 4x by 7 if you expand
this so you get this as the easy so we
have an example where we have to think a
little bit we are asked to find 98 to
the power 5 what we can do we have been
told that we have we know the formula of
a plus B to the power n or a minus B
power n so we know all this formula but
we don't know now I need to power 5 so
we can also say 98 is equal to hundred
minus 2 and 98 for 5 is going 200 minus
2 but fine and now using one binomial
theorem we can solve this so let's see
98 the power 5 is equal to 100 minus 2
is gone fine to expand this we'll get
this summation r is equal to 0 to 5 5 C
our build a hundred first hundred to the
power R into minus 2 to the power 5
minus R this is the formula we get for
NCR expansion what so now let's solve
this
so this become our is good 2 0 5 c 0 100
to the power 0 is equal to again 0 into
minus 2 power 5 plus 5 C 1 into hundred
to the power 1 and minus 2 to the power
4 plus 5 C 2 5c to 100th part 2 into
minus 2 to the power 3 plus 5 C 300 to
the power 3 into minus 2 to the power
this guy will decrease 2 plus 5 C 4 into
100 to the power 4 into minus 2 to the
power 1 plus 5 C 5 into 100 to the power
5 into minus to the power 0 this is the
expansion so expand this you get five
C's respond to one in the power 5 and to
the power Phi is equal to minus 32x
there so we'll say minus 32 and then Phi
C 1 is equal to 5 into 100 into minus 15
power 4 is equal to 16 plus 5 c 2 5
scene is good let's find out this year
we have done combination Phi C 2 is
equal to 5 permutation by 3 permutation
addition this bill to 5 into 4 into 3
formulation by 3 permutation into and
this is gone this is T equal to 10 so 5
C 2 is equal to 10 so I said n into 100
square that is 100 into hundred into
minus 2 whole cube is going to minus 8
this becomes a minus a minus a plus 5 C
3 also becomes same that is 10 into this
is 100 Q into minus 2 whole square is
equal to 4 plus 5 C 4 is equal to 1 into
102 the bar 4 2 2 3 4 in it 2 to the
power 1 into 2 minus 1 so we'll say
minus here and plus we see Phi C Phi is
equal to 1 10 to power 5 is equal to 3 4
5 times 2 0 into minus 2 to the power 0
is equal to 1 so you got this so if you
solve this what you get is 9:03 9:07 -
is the answer what we have done so we
have converted 90 to the power 5 into
100 - 2 / 5 we have applied the binomial
theorem and expanded this to get the
answer let's take one more example where
we think little more so we are asked to
tell which one is greater this artists
finding this power is difficult so we
can use the power of my limit we can
also write one point zero one is equal
to one plus point zero 1 to the power 1
2 3 1 by 16 so now I can expand this
expand this what we'll see is this guy
is nothing but 1 1 2 3 4 5 6 0 C naught
into 1 to the power 1 2 3 4 5 6 into
point 0 1 to the power 0 Plus this time
6 times this C 1 into 1 to the power 9
the subtract this 1 2 3 4 5 6 3 3 3 6 7
3 what is extra so you want this to get
smoother this is not there in 2.0 1/3
power 1 plus dollar when you solve this
100 C 0 is equal to 1 into 1 to the
power this cut the 1 into point zero 1
plus this C 1 this itself becomes this
much 1 2
or 6 into 1 to the power this is good to
one in 2.0 1 to the power 1 is going to
on by Amtrak plus again some constant
will say constant this also will receive
as a constant so this by this cancel so
this becomes 10,000 plus some constant
so that means this is always greater
than 10,000 so when we are supposed to
compare this and this so obviously this
is little because we have proved that 1
point 0 1 to the power 6 time this this
is nothing but n thousand plus some
constant so when he asked this on this
tenth of the mission is greater
obviously this greater because this is
equal to this plus embossing so then say
that this one is 3 so what we have done
we have written 1 point 0 1 to the power
n in the form of 1 plus 1 point 0 to the
power n we expanded this and we found
that this is nothing but 10,000 plus
constant that if this number is greater
than n thousand thank you visit exam
care.com
to watch free educational videos try
free online test if the best quality
study material study from the best tutor
and mentors and much forum times once
again
Related Exams
About this Video
|
4.64/5 Rating |
|
Nov 17, 2024 Last updated |
Video Description: Examples (NCERT) : Part 7 - Binomial Theorem for Airforce X Y / Indian Navy SSR 2024 is part of Mathematics for Airmen Group X preparation.
The notes and questions for Examples (NCERT) : Part 7 - Binomial Theorem have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus.
Information about Examples (NCERT) : Part 7 - Binomial Theorem covers all important topics for Airforce X Y / Indian Navy SSR 2024 Exam.
Find important definitions, questions, notes, meanings, examples, exercises and tests below for Examples (NCERT) : Part 7 - Binomial Theorem.
Introduction of Examples (NCERT) : Part 7 - Binomial Theorem in English is available as part of our Mathematics for Airmen Group X for Airforce X Y / Indian Navy SSR & Examples (NCERT) : Part 7 - Binomial Theorem in
Hindi for Mathematics for Airmen Group X course. Download more important topics related with notes, lectures and mock test series for
Airforce X Y / Indian Navy SSR Exam by signing up for free.
Description
Video Lecture & Questions for Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR - Airforce X Y / Indian Navy SSR full syllabus preparation | Free video for Airforce X Y / Indian Navy SSR exam to prepare for Mathematics for Airmen Group X.
Information about Examples (NCERT) : Part 7 - Binomial Theorem
Here you can find the meaning of Examples (NCERT) : Part 7 - Binomial Theorem defined & explained in the simplest way possible.
Besides explaining types of Examples (NCERT) : Part 7 - Binomial Theorem theory, EduRev gives you an ample number of questions to practice Examples (NCERT) : Part 7 - Binomial Theorem tests,
examples and also practice Airforce X Y / Indian Navy SSR tests.
|
149 videos|192 docs|197 tests
|
|
Explore Courses for Airforce X Y / Indian Navy SSR exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
video lectures
,Sample Paper
,Semester Notes
,Exam
,Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR
,Previous Year Questions with Solutions
,practice quizzes
,study material
,ppt
,Viva Questions
,Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR
,Extra Questions
,Examples (NCERT) : Part 7 - Binomial Theorem Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR
,mock tests for examination
,past year papers
,shortcuts and tricks
,MCQs
,Objective type Questions
,Summary
,Important questions
,Free
;
Study Examples (NCERT) : Part 7 - Binomial Theorem on the App
Students of Airforce X Y / Indian Navy SSR can study Examples (NCERT) : Part 7 - Binomial Theorem alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Examples (NCERT) : Part 7 - Binomial Theorem,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Examples (NCERT) : Part 7 - Binomial Theorem is prepared as per the latest Airforce X Y / Indian Navy SSR syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup