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Let $S = \sum_{i=3}^{100} i \log_{2} i$, and $T = \int_{2}^{100} x \log_{2}x dx$.

Which of the following statements is true?

  1. $S > T$
  2. $S = T$
  3. $S < T$ and $2S > T$
  4. $2S ≤ T$
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3 Answers

Best answer
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34 votes

$x\log_2 x$ is a continuously increasing function, and for a continuously increasing function $f(x)$,

$$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$$

But in question, summation of L.H.S. above, $a=3$ and in R.H.S, $a=2$, so we don't know whether $S > T$. So we compute some initial values :

$\sum_{x=3}^{4} x\log_2 x \approx 12.754$, and $\int_2^4 x\log_2 x = 11$

Since $\sum_{x=3}^{4} x\log_2 x > \int_2^4 x\log_2 x$, and since we already know that

$\sum_{x=5}^{100} x\log_2 x > \int_5^{100} x\log_2 x$

So $\sum_{x=3}^{100} x\log_2 x > \int_2^{100} x\log_2 x$

So S > T, and option (A) is correct.

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T=2log22+S

T=2+S

i.e. S<T

and

if 2S>T

2S>2+S

S>2  (that is true )

so 2S>T

and answer is C

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answer = option A

$x \log_2(x)$ is a continuously increasing function for the interval [2, 100]

We need to compare the numerical values 
S=$\sum_{i=3}^{100} i \log_{2} i$
T=$\int_{2}^{100} x \log_{2}x dx$

initially. 
$\sum_{x=3}^{4} x\log_2 x \approx 12.754$ and
$\int_2^4 x\log_2 x\ dx= 11$

we compute the actual values as:
S

 

T

where it is seen that S > T

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