its simple..the general term is r 10c r whole square so make one 10c r with r and convert it as 10 (9 choose r-1) and then you get two different series of (1+X)^n and expand and multiply and find coefficient of x^19
Very easy, differentiate (1+x)^n and multiply it with (1+x)^n, therefore the n-1 th term of the resultant binomial will give the resultant sum that is 10×19C9
Take 2 series (x+1)^10 = 10C0 x^10 + 10C1 x^9 + ...... + 10C10 .........(1) (1+x)^10 = 10C0 + 10C1 x + ...... + 10C10 x^10 .........(2) Differentiate 2 to get 10*(1+x)^9 = 10C1+2*10C2 x + 3* 10C3 x^2 + ........... 10 * 10C10 X^9 ...............(3) Multiply (1) and (3) and get the coefficient of x^9 from both sides. The right side is the expression given in the problem. The left side would be coefficient of X^9 in 10*(1+x)^19 which would be 10*19C9 which is same as 5 * 20C10
My solution: By differentiation of (x+1)^10 we can verify that, 1 * x^0 * 10C1 + ... + 10 * x^9 10C10 = 10(x+1)^9 ........(1) Hence multiply 10(x+1)^9 by (1+(1/x))^10. Now notice that the coefficient of 1/x is same as the desired sum[ Check this by expanding (1+(1/x))^10 and using (1)] But 10(x+1)^9 * (1+(1/x))^10 = 10((1+x)^19)/(x^10), hence coefficient of 1/x is clearly 10 * 19C9, and this is the desired answer.
u can use the fact that n choose k is the same as n choose (n-k) on the bigger than 5 terms, and than again the same fact to rewrite the sum as 5(10c0)² + 5(10c1)² + ... + 5(10c10)² which than simplifies to 5(20c10) its the same method as u used, just no need to add the sum again in the reversed order, so it makes it more intuitive
@@6Scarfy99 i accidentally discovered that (nc0)² + (nc1)² + ... + (ncn)² = (2ncn) while i was working on some problem another day. then i realized that i can use this fact here, i just had to get them to the same coefficient. after writing it like 10(10c0)² + 10(10c1)² + 10(10c2)² + 10(10c3)² + 10(10c4)² + 5(10c5)² it was clear that only 5 didn't have its pair, so i could divide others as 10(10c4)² = 5(10c4)² + 5(10c6)² for example come up with this by myself before watching his solution, but these two are very similar.
Write (10,r)²= (10,r)(10,10-r) Now let's us assume we have 10 boys and 10 girls. We need to form a team of 10 members in which there will be 1 girl captian always... The lhs depicts this.. Now let's us count this in another way... We need one girl who will be captain in 10c1 ways. We need 10-1=9 more members,who can be selected in (20-1,10-1) ways..total ways =10*19c9
Nice solution Here in India we have to learn the formula for these JEE ADVANCE problems I don't know how to type sigma or symbol for combination but anyway we are taught that sum from 0 to n of (nCr)^2=(2n)Cn
@Xpoz King paper me is type ka question solve nhi hoga. Bhai tujhe isliye easy laga to tune use fursat me solve kiya hoga ya tune uska solution dekh liya hoga. Vaise bhi do baar solve karne ya solutions dekne se to answer bhi yaad ho jata hai.
the sum ∑ⁿᵣ₌₀ (n r)² = (2n n) can be proven using vondermande's identity, that is ∑ʲᵣ₌₀ (n r)(m j-r) = (m+n j) Using combinatorial argument "Choosing j people from m men and n women in two ways, 1ˢᵗ method : (m+n j) 2ⁿᵈ method : calculating manually • j men, 0 woman = (m j)(n 0) • j-1 men, 1 woman = (m j-1)(n 1) • j-2 men, 2 women = (m j-2)(n 2) . . • 1 man, j-1 women = (m 1)(n j-1) • 0 man, j women = (m 0)(n j) adding all these, it is precisely the sum ∑ʲᵣ₌₀ (n r)(m j-r) thus 1ˢᵗ result = 2ⁿᵈ result" by setting m=n=j we have ∑ⁿᵣ₌₀ (n r)(n n-r) = (n+n n) applying the identity (n r) = (n n-r) ∑ⁿᵣ₌₀ (n r)(n r) = (2n n) ∑ⁿᵣ₌₀ (n r)² = (2n n)
its simple..the general term is r 10c r whole square so make one 10c r with r and convert it as 10 (9 choose r-1) and then you get two different series of (1+X)^n and expand and multiply and find coefficient of x^19
ua-cam.com/video/H1kQmnaVaTU/v-deo.html
This problem was so easy I don’t know why Indians keep complaining about the jee it seems easier than the SAT
@@dr.mikelitoris go and see full paper
@@dr.mikelitorisAlert: u will pee in ur pants after that
@@deepjyoti5610 ok I will
A challenging question and you made it simple to understand. Thank you
Very easy, differentiate (1+x)^n and multiply it with (1+x)^n, therefore the n-1 th term of the resultant binomial will give the resultant sum that is 10×19C9
Not with calculus plz just pure aljebra
Take 2 series
(x+1)^10 = 10C0 x^10 + 10C1 x^9 + ...... + 10C10 .........(1)
(1+x)^10 = 10C0 + 10C1 x + ...... + 10C10 x^10 .........(2)
Differentiate 2 to get
10*(1+x)^9 = 10C1+2*10C2 x + 3* 10C3 x^2 + ........... 10 * 10C10 X^9 ...............(3)
Multiply (1) and (3) and get the coefficient of x^9 from both sides.
The right side is the expression given in the problem. The left side would be coefficient of X^9 in 10*(1+x)^19 which would be 10*19C9 which is same as 5 * 20C10
Upload more problems of jee advance
Lol these are all basic level questions but its to tough to solve but u solve like a genius
My solution:
By differentiation of (x+1)^10 we can verify that,
1 * x^0 * 10C1 + ... + 10 * x^9 10C10 = 10(x+1)^9 ........(1)
Hence multiply 10(x+1)^9 by (1+(1/x))^10. Now notice that the coefficient of 1/x is same as the desired sum[ Check this by expanding (1+(1/x))^10 and using (1)]
But 10(x+1)^9 * (1+(1/x))^10 = 10((1+x)^19)/(x^10), hence coefficient of 1/x is clearly 10 * 19C9, and this is the desired answer.
u can use the fact that n choose k is the same as n choose (n-k) on the bigger than 5 terms, and than again the same fact to rewrite the sum as
5(10c0)² + 5(10c1)² + ... + 5(10c10)²
which than simplifies to 5(20c10)
its the same method as u used, just no need to add the sum again in the reversed order, so it makes it more intuitive
Great method mate , did you realise it after seeing the answer ? Or was this your first thought?
@@6Scarfy99 i accidentally discovered that (nc0)² + (nc1)² + ... + (ncn)² = (2ncn) while i was working on some problem another day. then i realized that i can use this fact here, i just had to get them to the same coefficient. after writing it like 10(10c0)² + 10(10c1)² + 10(10c2)² + 10(10c3)² + 10(10c4)² + 5(10c5)²
it was clear that only 5 didn't have its pair, so i could divide others as
10(10c4)² = 5(10c4)² + 5(10c6)² for example
come up with this by myself before watching his solution, but these two are very similar.
Write (10,r)²= (10,r)(10,10-r)
Now let's us assume we have 10 boys and 10 girls.
We need to form a team of 10 members in which there will be 1 girl captian always...
The lhs depicts this..
Now let's us count this in another way...
We need one girl who will be captain in 10c1 ways.
We need 10-1=9 more members,who can be selected in (20-1,10-1) ways..total ways =10*19c9
Nice analogy.
Nais
Please make a playlist of calculus related problems.
Nice solution! Keep your great work!
From Indonesia?
@@wise_math yea i am
@@fedryfirman.a5783 olympiad student?
idk what it is called, but kind of 😁
Nice solution
Here in India we have to learn the formula for these JEE ADVANCE problems
I don't know how to type sigma or symbol for combination but anyway
we are taught that sum from 0 to n of (nCr)^2=(2n)Cn
No in india , probably in the school you go. But If you go to a good coaching , they will teach you 3-4 different methods for this same problem
Brother haven't you been taught the method of multiplication of 2 binomial expansions????
Don't put your condition on everyone
I know how to derive it
Thank you so much, I love this problem and the solution.
Whenever coefficient are in AP just reverse the series and add and it will give fruitful result
no this not an agp that trick will not work here
@@prathamrathore776 pratham to phir video me kya Kiya h sir ne 😂😂 (and I meant coefficients i.e. 1,2,3,4...10 in Ap)
@@aniketmehta4104 bro see carefully he didnt add but multiplied the terms of two diff expansions
@@prathamrathore776 or in other words added the values of coefficients by taking the (10 n) common where n = 1,2,3,…,10
Keep posting mate
Where does the 1430 come from?
It's in the question
Wowww!! TY
more pls
Actually it could be done by some shortcut which I know...My teacher taught
WOW!
Bro just take the problem no :12 of paper 2 from jee advanced -2020 of binomial theorm.
@Xpoz King yo bro easy hai
@Xpoz King paper me is type ka question solve nhi hoga. Bhai tujhe isliye easy laga to tune use fursat me solve kiya hoga ya tune uska solution dekh liya hoga. Vaise bhi do baar solve karne ya solutions dekne se to answer bhi yaad ho jata hai.
good brother
voice is not legible!!! is it understood by the narrator?
I think your explanation is too long. You should go faster. I have almost felt asleep.
the sum ∑ⁿᵣ₌₀ (n r)² = (2n n)
can be proven using vondermande's identity, that is
∑ʲᵣ₌₀ (n r)(m j-r) = (m+n j)
Using combinatorial argument
"Choosing j people from m men and n women in two ways,
1ˢᵗ method : (m+n j)
2ⁿᵈ method : calculating manually
• j men, 0 woman = (m j)(n 0)
• j-1 men, 1 woman = (m j-1)(n 1)
• j-2 men, 2 women = (m j-2)(n 2)
.
.
• 1 man, j-1 women = (m 1)(n j-1)
• 0 man, j women = (m 0)(n j)
adding all these, it is precisely the sum
∑ʲᵣ₌₀ (n r)(m j-r)
thus 1ˢᵗ result = 2ⁿᵈ result"
by setting m=n=j
we have
∑ⁿᵣ₌₀ (n r)(n n-r) = (n+n n)
applying the identity (n r) = (n n-r)
∑ⁿᵣ₌₀ (n r)(n r) = (2n n)
∑ⁿᵣ₌₀ (n r)² = (2n n)