- r - r -1 nCrn -A- n -2 n-1B- nC -2 nD- 2 n

The correct option is A n.2n 11+xn nbsp;=n0+n1x+n2x2+....+nnxndifferentiating nbsp;w.r.to nbsp;xn.1+xn 1= nbsp;n1+2xn2+....+n.xn 1nnput nbsp;x=1n.2n 1 ...

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

Byju's Answer

Standard XII

Mathematics

Differentiation to Solve Modified Sum of Binomial Coefficients

∑ r · r =1 nC...

Question

$n \sum r \cdot r = 1 n C r =$

n . 2^{n - 1}

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

n

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

2 n

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

2^{n}

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

Open in App

Solution

The correct option is A $n . 2^{n - 1}$
${(1 + x)}^{n} = (\begin{matrix} n 0 \end{matrix}) + (\begin{matrix} n 1 \end{matrix}) x + (\begin{matrix} n 2 \end{matrix}) x^{2} + . . . . + (\begin{matrix} n n \end{matrix}) x^{n} differentiating w . r . to x n . {(1 + x)}^{n - 1} = (\begin{matrix} n 1 \end{matrix}) + 2 x (\begin{matrix} n 2 \end{matrix}) + . . . . + n . x^{n - 1} (\begin{matrix} n n \end{matrix}) put x = 1 n . 2^{n - 1} = (\begin{matrix} n 1 \end{matrix}) + 2 (\begin{matrix} n 2 \end{matrix}) + . . . . + n . (\begin{matrix} n n \end{matrix}) n . 2^{n - 1} = n \sum r = 1 r \cdot n C r hence n . 2^{n - 1} is the answer .$

Suggest Corrections

Similar questions

If $C_{r}$ means $^{n} C_{r}$ then $\frac{C_{0}}{1} + \frac{C_{2}}{3} + \frac{C_{4}}{5} + \dots =$

Q. If

n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

, and

n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}

, then

p

is equal to

Find a pattern for these sequences:

a = 1, 2, 3, 4...
b = 0, 2, 4, 6, 8...
c = 1, 3, 5, 7...

\sum_{r = 1}^{n} n C r

The sum of the first $n$ terms of the series
$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to ___.

Join BYJU'S Learning Program

n∑r⋅r=1nCr=

The correct option is A n.2n−1(1+x)n =(n0)+(n1)x+(n2)x2+....+(nn)xndifferentiating w.r.to xn.(1+x)n−1= (n1)+2x(n2)+....+n.xn−1(nn)put x=1n.2n−1 = (n1)+2(n2)+....+n.(nn)n.2n−1 = n∑r=1r⋅nCr hence n.2n−1 is the answer.

$n \sum r \cdot r = 1 n C r =$