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Tarunvir Singh
Subject: Maths
, asked on 11/12/17
Do question 1
Answer
1
Pavithra Saravanan
Subject: Maths
, asked on 28/11/17
If the fourth term in the expansion of (ax+1/x)n is 20/27 then find the value of n and a . Also find its 4th term from the end
Answer
1
Shivam Desale
Subject: Maths
, asked on 14/11/17
Qn no. 5
Answer
2
Soumyadeep Kar
Subject: Maths
, asked on 12/11/17
Q. Prove that
$\sum _{r=1}^{k}{\left(-3\right)}^{r-13n}{C}_{2r-1}=0,whereK=\frac{3n}{2}$
and n is an even positive integer.
Answer
1
Soumyadeep Kar
Subject: Maths
, asked on 12/11/17
Q. Find the sum of series
$\frac{3}{1!}+\frac{5}{2!}+\frac{9}{3!}+\frac{15}{4!}+\frac{23}{5!}+..................\infty .$
Answer
3
Shivam Desale
Subject: Maths
, asked on 10/11/17
Qn no. 1
Answer
1
Nura Mohammad
Subject: Maths
, asked on 9/11/17
Which of the following is larger
99^50+100^50 or 101^50
The part of the answer that is given in the photo please explain. How can we take 100^50 common?
$=2\left[{}^{50}C_{1}\right(100{)}^{49}+{}^{50}C_{3}(100{)}^{47}+.....]\phantom{\rule{0ex}{0ex}}=2\left[50\right(100{)}^{49}+{}^{50}C_{3}(100{)}^{47}+....]\phantom{\rule{0ex}{0ex}}={100}^{50}+otherpositiveterms\phantom{\rule{0ex}{0ex}}{100}^{50}$
$So,{101}^{50}-{99}^{50}>{100}^{50},i.e.,{101}^{50}>{99}^{50}+{100}^{50}\phantom{\rule{0ex}{0ex}}Hence,{101}^{50}islargerthan{99}^{50}+{100}^{50}$
Answer
1
Anushka Agrawal
Subject: Maths
, asked on 3/11/17
Please tell me the mistake in my answer to the following question as the answer is r=4
Answer
1
Soumyadeep Kar
Subject: Maths
, asked on 2/11/17
Find the coefficient of x
^{3}
in the expansion of (1-2x+3x
^{2}
-4x
^{3}
)
Answer
3
Soumyadeep Kar
Subject: Maths
, asked on 2/11/17
Solve this:
Q). Find the coefficient of
${x}^{2}{y}^{4}{z}^{2}$
in the expansion of
${\left(2x-3y+4z\right)}^{9}$
.
Answer
1
Shivam Gupta
Subject: Maths
, asked on 2/11/17
sir
Answer
1
Shivam Gupta
Subject: Maths
, asked on 2/11/17
1
sir
Answer
1
Shivam Gupta
Subject: Maths
, asked on 2/11/17
sir
Answer
1
Shivam Gupta
Subject: Maths
, asked on 2/11/17
sir
Answer
1
Shivam Gupta
Subject: Maths
, asked on 2/11/17
sir
Answer
1
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99^50+100^50 or 101^50

The part of the answer that is given in the photo please explain. How can we take 100^50 common?

$=2\left[{}^{50}C_{1}\right(100{)}^{49}+{}^{50}C_{3}(100{)}^{47}+.....]\phantom{\rule{0ex}{0ex}}=2\left[50\right(100{)}^{49}+{}^{50}C_{3}(100{)}^{47}+....]\phantom{\rule{0ex}{0ex}}={100}^{50}+otherpositiveterms\phantom{\rule{0ex}{0ex}}{100}^{50}$

$So,{101}^{50}-{99}^{50}>{100}^{50},i.e.,{101}^{50}>{99}^{50}+{100}^{50}\phantom{\rule{0ex}{0ex}}Hence,{101}^{50}islargerthan{99}^{50}+{100}^{50}$

^{3}in the expansion of (1-2x+3x^{2}-4x^{3})Q). Find the coefficient of ${x}^{2}{y}^{4}{z}^{2}$ in the expansion of ${\left(2x-3y+4z\right)}^{9}$.