The value of 12-n C1-22-n C2-32-n C3-n2-n Cn isA- nn-1 - 2n-1B- nn-1 - 2n-2C- n n -1 - 2 n -2D- n n -1 - 2 n -1

The correct option is C n(n+1).2^n 21^2. ^nC_1+2^2. ^nC_2+3^2. ^nC_3+...+n^2. ^nC_n=∑^n_r=1(r^2) ^nC_r Now ∑^n_r=1(r^2) ^nC_r=∑^n_r=1{r(r 1)+r}^nC_r =∑^n_r=1r ...

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

Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

The value of ...

Question

The value of
$1^{2} .^{n} C_{1} + 2^{2} .^{n} C_{2} + 3^{2} .^{n} C_{3} + . . . + n^{2} .^{n} C_{n}$ is

n (n + 1) {.2}^{n - 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n (n + 1) {.2}^{n - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n (n - 1) {.2}^{n - 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n (n - 1) {.2}^{n - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $n (n + 1) {.2}^{n - 2}$
$1^{2} .^{n} C_{1} + 2^{2} .^{n} C_{2} + 3^{2} .^{n} C_{3} + . . . + n^{2} .^{n} C_{n} = n \sum r = 1 (r^{2})^{n} C_{r} Now n \sum r = 1 (r^{2})^{n} C_{r} = n \sum r = 1 {r (r - 1) + r}^{n} C_{r} = n \sum r = 1 r (r - 1) .^{n} C_{r} + n \sum r = 1 r .^{n} C_{r} = n \sum r = 2 n (n - 1) .^{n - 2} C_{r - 2} + n \sum r = 1 n .^{n - 1} C_{r - 1} (r/ . \frac{n (n - 1)!}{r/ (n - r)! (r - 1)!}) = n (n - 1) n \sum r = 2^{n - 2} C_{r - 2} + n n \sum r = 1^{n - 1} C_{r - 1} = n (n - 1) {.2}^{n - 2} + n {.2}^{n - 1} = n {.2}^{n - 2} (n - 1 + 2) = n (n + 1) {.2}^{n - 2}$

Suggest Corrections

Similar questions

Q. The mean of the observations

0, 1, 2, 3, \dots, n

having corresponding weight's

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

respectively is:

Q. The mean of the values

0, 1, 2, \dots \dots \dots, n

having corresponding weight

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots \dots \dots \dots^{n} C_{n},

respectively is

If $^{m} C_{1} =^{n} C_{2},$ then

$I f f : (0, \infty) \to (0, \infty)$ satisfy

$f (x f (y)) = x^{2} y^{2} (a \in R)$ ,then

$\sum_{r = 1}^{n} - f (r)^{n} C_{r} i s$

Q. Assertion :

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

because Reason:

\frac{1}{(1) (3)} + \frac{1}{(3) (5)} + . . . . . + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2 n + 1}

Join BYJU'S Learning Program

The value of 12.nC1+22.nC2+32.nC3+...+n2.nCn is

The value of
$1^{2} .^{n} C_{1} + 2^{2} .^{n} C_{2} + 3^{2} .^{n} C_{3} + . . . + n^{2} .^{n} C_{n}$ is