S=∑r=02018(−1)r 2018Cr1+rln2(1+ln22018)r then [S] equal to (where [.] is G.I.F.)

S=∑r=0n(−1)rnCr1+rx(1+nx)r(let n = 2018, x = ln 2)=∑r=0n(−1)r nCr(1+nx)r+∑r=0n(−1)r.nr.n−1Cr−1rx(1+nx)r =∑r=0n(−1)r nCr(1+nx)r−nx(1+nx)∑r=0n(−1)r−ln−1Cr−11(1+nx)r−1 =(1−11+nx)n−nx1+nx(1−11+nx)n−1 =(nx1+nx)n−nx1+nx(nx1+nx)n−1=0

S=∑r=02018(−1)r 2018Cr1+rln2(1+ln22018)r then [S] equal to (where [.] is G.I.F.)

S=∑r=0n(−1)rnCr1+rx(1+nx)r(let n = 2018, x = ln 2)=∑r=0n(−1)r nCr(1+nx)r+∑r=0n(−1)r.nr.n−1Cr−1rx(1+nx)r =∑r=0n(−1)r nCr(1+nx)r−nx(1+nx)∑r=0n(−1)r−ln−1Cr−11(1+nx)r−1 =(1−11+nx)n−nx1+nx(1−11+nx)n−1 =(nx1+nx)n−nx1+nx(nx1+nx)n−1=0

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$S = \sum_{r = 0}^{2018} {(- 1)}^{r}^{2018} C_{r} \frac{1 + r \ln 2}{{(1 + \ln 2^{2018})}^{r}}$ then [S] equal to (where [.] is G.I.F.)

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Number of non-negative solutions of equation $\sqrt{l n (- x)} = l n | x |$

– 4

answer is B, D.

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Detailed Solution

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

(let n = 2018, x = ln 2)

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