Statement -1: _{r=0}ⁿ (r+1)ⁿ {}ⁿ C_r = (n+2) 2ⁿ-1}. Statement - 2: _{r=0}ⁿ (r+1)ⁿ {}ⁿ C_r x^r = (1+x)ⁿ + n x (1+x)ⁿ-1}. [AIEEE 2008]

Correct Answer is : Statement −1-1−1 is true, Statement −2-2−2 is true; Statement −2-2−2 is a correct explanation for Statement - 1

Statement -1 is false, Statement - 2 is true

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

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

Statement -1 is true, Statement -2 is false

Correct Answer is : Statement −1-1−1 is true, Statement −2-2−2 is true; Statement −2-2−2 is a correct explanation for Statement - 1

Test Index

Binomial Theorem Part 1

Section: Mathematics

Question : 82 of 100

Marks: +1, -0

Statement

-1: \sum\limits_{r=0}^{n} (r+1)^n {}^n C_r = (n+2) 2^{n-1}

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

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

Solution:

Check Statement - 1

\sum\limits_{r=0}^{n} (r+1) {}^n C_r

=\sum\limits_{r=0}^{n} r \cdot {}^n C_r + \sum\limits_{r=0}^{n} {}^n C_r

=\sum\limits_{r=1}^{n} r \cdot \frac{n}{r} {}^{n-1}C_{r-1} + 2^n

=n \sum\limits_{r=1}^{n} {}^{n-1}C_{r-1} + 2^n

=n \times 2^{n-1} + 2^n

=2^{n-1}[n+2]

Check Statement 2 :

\sum\limits_{r=0}^{n} (r+1)^n {}^n C_r x^r

=\sum\limits_{r=0}^{n} r \cdot {}^n C_r \cdot x^r + \sum\limits_{r=0}^{n} {}^n C_r \cdot x^r

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

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

=n x (1+x)^{n-1} + (1+x)^n

Substitude

x=1

in the statement 2 and we get,

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

Go to Question:

Prev Question

Next Question