Let fn- k-1 n k 2 C k 2 find f5

f(n)=∑ _ k=1 ^ n k^ 2 (C_ r )^ 2 k( C _ k )=k n! / k!(n k)! = n(n 1)! (k 1)!(n k)! =n( _ nbsp;^ n 1 C _ k 1 ) ∑ _ k=1 ^ n k ^ 2 C _ k ^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

Let fn=∑ k=...

Question

Let $f (n) = \sum_{k = 1}^{n} k^{2} C_{k}^{2}$ find $f (5)$

Open in App

Solution

$f (n) = \sum_{k = 1}^{n} k^{2} (C_{r})^{2} k (C_{k}) = k \frac{n!}{k! (n - k)!} = \frac{n (n - 1)!}{(k - 1)! (n - k)!} = n ({^{n - 1} C}_{k - 1}) \sum_{k = 1}^{n} k^{2} C_{k}^{2} = \sum_{k = 1}^{n} {(k \frac{n!}{k! (n - k)!})}^{2} = \sum_{k = 1}^{n} (n ({^{n - 1} C}_{k - 1}))^{2} = n^{2} \sum_{k = 1}^{n} {({^{n - 1} C}_{k - 1})}^{2} f (n) = n^{2} \sum_{k = 1}^{n} {({^{n - 1} C}_{k - 1})}^{2} (1) W e k n o w t h a t (1 + x)^{n} = C_{0} + C_{1} x^{1} + C_{2} x^{2} + . . . . . . . . C_{n} x^{n} (1 + x)^{n} = C_{0} x^{n} + C_{1} x^{n - 1} + C_{2} x^{n - 2} + . . . . . . . . C_{n} M u l t i p l y B o t h t h e e q u a t i o n s a n d C o m p a r e t h e c o e f f i c i e n t s o f x^{n} W e g e t^{2 n} C_{n} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . . . C_{n}^{2} W h e n n = n - 1 t h e n^{2 n - 2} C_{n - 1} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . . . C_{n - 1}^{2} (2) F r o m (1) w e h a v e f (n) = n^{2} (C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . . . C_{n - 1}^{2}) U s i n g (2) W e g e t f (n) = (n^{2})^{2 n - 2} C_{n - 1} f (5) = (5^{2})^{10 - 2} C_{5 - 1} = 25^{8} C_{4} = 1750$

Suggest Corrections

Similar questions

Q. Let

f (n) = n \sum r = 1 r^{2} {(^{n} C_{r})}^{2}

then the value of

f (5)

is equal to

Let $f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 1} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 1} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Then, f(n) is divisible by

Q. Let

f (n) = n \sum k = 1 {cot}^{- 1} (k^{2} + k + 1) .

Then which of the following options is CORRECT ?

Q. If

F (n) = (- 1)^{k - 1} (n - 1), G (n) = n - F (n)

then

(G o G) (n) =

(where

k

is odd)

Q. Let

f (1) = 1

and

f (n) = 2 \sum_{r = 1}^{n - 1} f (r)

. Then

\sum_{n = 1}^{m} f (n)

. is equal to

Join BYJU'S Learning Program

Let f(n)=∑nk=1k2C2k find f(5)

Let $f (n) = \sum_{k = 1}^{n} k^{2} C_{k}^{2}$ find $f (5)$