Prove that the coefficient of (r+1)th term in the expansion of $(1 + x)^{n + 1}$ is equal to the sum of the coefficients of rth and (r+1)th terms in the expansion of $(1 + x)^{n} .$

Q. Assertion :Let

f (x) = x^{n}

f^{'} (x) = r!^{n} C_{r} x^{n - r}

denotes the

r^{t h}

order derivative of

f (x)

then

f (1) + \frac{f^{'} (1)}{1!} + \frac{f^{''} (1)}{2!} + . . . . . + \frac{f^{n} (1)}{n!} = 2^{n}

Reason: The sum of binomial coefficients in the expansion of

{(1 + x)}^{n}

2^{n}

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What is the rth term in the expansion of a binomial (x+y)n?

The correct option is B nCr−1.xn−r+1.yr−1Tr+1=nCrxn−ryrPut r=r−1⇒Tr−1+1=nCr−1xn−(r−1)yr−1⇒Tr=nCr−1xn−r+1yr−1So, option A is correct.

What is the $r^{t h}$ term in the expansion of a binomial $(x + y)^{n}$ ?

The correct option is B $^{n} C_{r - 1} . x^{n - r + 1} . y^{r - 1}$
$T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$
Put $r = r - 1$
$\Rightarrow T_{r - 1 + 1} =^{n} C_{r - 1} x^{n - (r - 1)} y^{r - 1} \Rightarrow T_{r} =^{n} C_{r - 1} x^{n - r + 1} y^{r - 1}$
So, option A is correct.