If (1+x)n =C0+C1x+C2x2+....Cnxn, then ∑r=0n∑s=0n(r+s)CrCs is equal to :

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If (1+x)^n = C_0 + C_1 x + C_2 x^2 + .... C_nx^n , then ^n_{r

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... C_{n} x^{n},$ then $\sum_{r = 0}^{n} \sum_{s = 0}^{n} (r + s) C_{r} C_{s}$ is equal to :

2^{2 n}

n . 2^{2 n}

n . 2^{n}

None of these

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William Discussed

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... C_{n} x^{n},

then

\sum_{r = 0}^{n} \sum_{s = 0}^{n} (r + s) C_{r} C_{s}

is equal to :

14 mins ago

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14 mins ago

I = r = 0 \sum n s = 0 \sum n (r + s) C_{r} C_{s}

= r = 0 \sum n s = 0 \sum n^{r} C_{r} C_{s} + r = 0 \sum n s = 0 \sum n C_{r} C_{s}

I = r = 0 \sum n^{r} C_{r} . s = 0 \sum n C_{3} +^{n} s = 0 \sum S C_{s} . r = 0 \sum n C_{r}

(A)

A = r = 0 \sum n^{r} C_{r} . s = 0 \sum n c^{s}

(1 + x)^{n} = C_{0} + C_{1} x + ...

Put

x + 1

2^{n} = r = 0 \sum n C_{s}

...(1)

(1 + x)^{n} = C_{0} + C_{1} + x ..... C_{n} x^{n}

differentiate with respect to

x

n (1 + x)^{n - 1} = C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + ....

Put

x = 1

n (2)^{n - 1} = r = 0 \sum n^{r} C r

(B)

A = r = 0 \sum n^{r} C_{r} . s = 0 \sum n C_{5}

A = n 2^{n - 1} . 2^{n}

from(1) and (2)

B = s = 0 \sum n^{s} C_{n} . r = 0 \sum n C_{r}

B = n 2^{n - 1} . 2^{n}

from(1) and (2)

I = (A + B - n . 2^{n} . 2^{n - 1}) .2

I = n . 2^{2 n}

∴ I + n . 2^{2 n}

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Question Text	If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... C_{n} x^{n},$ then $\sum_{r = 0}^{n} \sum_{s = 0}^{n} (r + s) C_{r} C_{s}$ is equal to :
Answer Type	Text solution:1
Upvotes	150