Statement-1- -r-0nr-1 nCr - n-22n-1-Statement-2- -r-0nr-1 nCr xr - 1-xn -nx1-xn-1-

We have ∑_r=0^n (r+1)^nC_rx^r =∑_r=0^n r^nC_rx^r+ ∑_r=0^n^nC_rx^r = ∑_r=1^n r.n/r^n 1C_r 1x^r + (1+x)^n = nx ∑_r=1^n^n 1 C_r 1x^r 1 + (1+x)^n =nx(1+x)^n 1+( ...

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Byju's Answer

Standard XIII

Mathematics

Differentiation to Solve Modified Sum of Binomial Coefficients

Statement-1: ...

Question

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .$
Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .$

Statemet -1 is false, Statement - 2 is ture

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Statemet -1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

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Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1

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Statement -1 is true, Statement -2 is false

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Solution

The correct option is B Statemet -1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
We have
$\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = \sum_{r = 0}^{n} r^{n} C_{r} x^{r} + \sum_{r = 0}^{n}^{n} C_{r} x^{r} = \sum_{r = 1}^{n} r . {\frac{n}{r}}^{n - 1} C_{r - 1} x^{r} + (1 + x)^{n} = n x \sum_{r = 1}^{n}^{n - 1} C_{r - 1} x^{r - 1} + (1 + x)^{n} = n x (1 + x)^{n - 1} + (1 + x)^{n} =$ RHS
$∴$ Statement 2 is correct.
Putting x=1, we get
$\sum_{r = 0}^{n} (r + 1)^{n} c_{r} = n . 2^{n - 1} + 2^{n} = (n + 2) {.2}^{n - 1}$
$∴$ Statement 1 is also true and statement 2 is a correct explanation for statement 1.

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Similar questions

Q. Statement-l:

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

Statement-2:

n \sum r = 0 (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}

Q. Statement-1:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

Statement-2:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

Q. Statement-1:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

Statement-2:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

Q. Assertion :

\sum_{r = 0}^{n} {r + 1}^{n} C_{r} = (n + 2) 2^{n - 1}

Reason:

\sum_{r = 0}^{n} {r + 1}^{n} C_{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}

Q. If

n \sum r = 0 (\frac{r + 2}{r + 1})^{n} C_{r} = \frac{2^{8} - 1}{6}

, then

n =

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Statement-1: ∑nr=0(r+1)nCr=(n+2)2n−1. Statement-2: ∑nr=0(r+1)nCrxr=(1+x)n+nx(1+x)n−1.

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .$
Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .$