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Algebra of Derivatives

If S n deno...

Question

If $S_{n}$ denotes the sum of $n$ terms of $g . p$ . whose common ratio is $r$ , then $(r - 1) \frac{d S_{n}}{d r}$ is equal to

A

(n - 1) S_{n} + n S_{n - 1}

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B

(n - 1) S_{n} - n S_{n - 1}

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C

(n - 1) S_{n}

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D

None of these

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Solution

The correct option is D $(n - 1) S_{n} - n S_{n - 1}$
We have, $S_{n} = \frac{a (r^{n} + 1)}{r - 1}$
$\Rightarrow (r - 1) S_{n} = {a r}^{n} - a$
Differentiating both sides with respect to $r,$ we get
$(r - 1) \frac{{d S}_{n}}{d r} + S_{n} = n a r^{n - 1} - 0$
$\Rightarrow (r - 1) \frac{{d S}_{n}}{d r} = n a r^{n - 1} - S_{n}$
$= n [n$ th term from of $G . P .]$ $- S_{n}$
$= n (S_{n} - S_{n - 1}) - S_{n}$
$= (n - 1) S_{n} - n S_{n - 1} .$

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Similar questions

Q. Assertion :If

S_{n}

denotes the sum of

n

terms a series given by

S_{n} = \frac{n (n + 1) (n + 2)}{6} \forall n \geq 1,

lim n \to \infty n \sum r = 1 \frac{1}{t_{r}} = 4

t_{n} = S_{n} - S_{n - 1}

Q. Assertion :Let

f (n) = n \sum r = 1 r^{4}, t h e n n \sum r = 1 r {(n - r)}^{3}

n {(\frac{n (n + 1)}{2})}^{2} - f (n)

n \sum r = 1 r {(n - 1)}^{3} = n \sum r = 1 (n - 1) r^{3}

C_{r}

is coefficient of

x^{r}

in expansion of

{(1 + x)}^{n}, n ϵ N, n \geq 4

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}}

is divisible (remainder 0) by
(1+x)

^{n}

ϵ N

\geq 4

c_{r}, x^{r}

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}} = 0

Q. Assertion :Let

f (x) = x^{n}

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

r^{t h}

order derivative of

f (x)

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}

Q. Assertion :If

f (x) = x^{n}

f (1) + \frac{f^{'} (1)}{1} + \frac{f^{''} (1)}{2!} + \frac{f^{'''} (1)}{3!} + . . . + \frac{^{n} f (1)}{n!} = 2^{n}

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

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

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