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Stuti Srivastava
Subject: Math
, asked on 20/5/18
How to break (k^3 +6k^2+9k+4) this such that it becomes (k+1)(k+1)(k+4).
Answer
1
Sanitya Srivastava
Subject: Math
, asked on 28/4/18
Experts kindly help
Answer
3
Saswat Das
Subject: Math
, asked on 14/4/18
Q.
Find the differentiation
$1.y={\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2x}\phantom{\rule{0ex}{0ex}}2.y={\left({\mathrm{cos}}^{2}x+1\right)}^{3}\phantom{\rule{0ex}{0ex}}3.y=\mathrm{cos}{x}^{3}+{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}4.y=\frac{{\left(3{x}^{2}-9\right)}^{2}}{\mathrm{sin}x}$
Answer
1
Drishti
Subject: Math
, asked on 23/3/18
$Showthat:-\phantom{\rule{0ex}{0ex}}\left|\begin{array}{ccc}1& 1& 1+3x\\ 1+3y& 1& 1\\ 1& 1+3x& 1\end{array}\right|=9(3xyz+xy+yx+zx)$
Answer
2
Ridhit Jain
Subject: Math
, asked on 6/3/18
Prove by PMI
$3.{2}^{2}+{3}^{2}.{2}^{3}+...........+{3}^{n}{2}^{n+1}=\frac{12}{5}\left({6}^{n}-1\right)$
prove
Answer
1
Kanishk Sharma
Subject: Math
, asked on 27/2/18
Q.4. Prove that
${3}^{3n}-26n-1$
is divisible by 676.
Answer
1
Dharunika Vijayakumar
Subject: Math
, asked on 14/2/18
TELL ME THE ANSWER FOR 15th QUESTION . PLEASE TELL ME FAST
Q15. Prove that ( cos
$\theta +i\mathrm{sin}\theta $
)
^{n}
= cos
$\left(\mathrm{n}\mathrm{\theta}\right)+i\mathrm{sin}(\mathrm{n}\mathrm{\theta})$
, for all n
$\in $
N by using PMI.
Answer
3
Tanmaya Darisi
Subject: Math
, asked on 8/2/18
Using PMI prove that 1 × 1! + 2 × 2! + 3 × 3! + -----+ n × n! = (n + 1)! – 1 for all n∈N
Answer
1
Naga Nandhini
Subject: Math
, asked on 31/1/18
Experts,explain the one underlined after
Answer
1
Naga Nandhini
Subject: Math
, asked on 31/1/18
prove that 1+2+3+....+n=n(n+1)/2 by pmi
Answer
1
Ashwini Upadhya
Subject: Math
, asked on 20/1/18
Please solve it using mathematical induction
$Provethat(1+x{)}^{n}\ge \left(1+nx\right),forallnaturalnumbern,wherex-1bymathematicalinduction.$
Answer
2
Ranjan Shivam
Subject: Math
, asked on 2/1/18
Q. if a, b, c, are in A.P. , then
$\frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}{\displaystyle +}{\displaystyle \sqrt{a}}},\frac{1}{\sqrt{a}{\displaystyle +}{\displaystyle \sqrt{b}}}arein\phantom{\rule{0ex}{0ex}}\left(A\right)A.P.\left(B\right)G.P.\phantom{\rule{0ex}{0ex}}\left(C\right)H.P.\left(D\right)Noneofthese$
Answer
1
Nura Mohammad
Subject: Math
, asked on 25/11/17
Q. Prove that
${2}^{n}>nforallpositiveintegersn.$
Please explain the last line of solution
Answer
1
Nura Mohammad
Subject: Math
, asked on 25/11/17
How is the underlined step coming?
Answer
1
Nura Mohammad
Subject: Math
, asked on 25/11/17
Prove the following by the principle of mathematical induction.
2n?>?n2, where?n?is a positive integer such that?n?> 4.
Solution:
Let the given statement be P(n), i.e.,
P(n) : 2n?>?n2?where?n?> 4
For?n?= 5,
25?= 32 and 52?= 25
?25?> 52
Thus, P(n) is true for?n?= 5.
Let P(n) be true for?n?=?k, i.e.,
2k?>?k2?? (1)
Now, we have to prove that P(k? 1) is true whenever P(k) is true, i.e. we have to prove that 2k? 1?> (k? 1)2.
From equation (1), we obtain
2k?>?k2
On multiplying both sides with 2, we obtain
2 ? 2k?> 2 ??k2
2k? 1?> 2k2
?To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2.
Let us assume 2k2?> (k? 1)2.
? 2k2?>?k2? 2k? 1
??k2?> 2k? 1
??k2?? 2k?? 1 > 0
? (k?? 1)2?? 2 > 0
? (k?? 1)2?> 2, which is true as?k?> 4
Hence, our assumption 2k2?> (k? 1)2?is correct and we have 2k? 1?> (k? 1)2.
Thus, P(n) is true for?n?=?k? 1.
Thus, by the principle of mathematical induction, the given mathematical statement is true for every positive integer?n.
?
IN THIS EXAMPLE IT HAS BEEN WRITTEN THAT ,
To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2.
how?2k? 1?> (k? 1)2?=?2k2?> (k? 1)2
I cant understand how so plz explain that
Answer
1
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Q. Find the differentiation

$1.y={\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2x}\phantom{\rule{0ex}{0ex}}2.y={\left({\mathrm{cos}}^{2}x+1\right)}^{3}\phantom{\rule{0ex}{0ex}}3.y=\mathrm{cos}{x}^{3}+{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}4.y=\frac{{\left(3{x}^{2}-9\right)}^{2}}{\mathrm{sin}x}$

$3.{2}^{2}+{3}^{2}.{2}^{3}+...........+{3}^{n}{2}^{n+1}=\frac{12}{5}\left({6}^{n}-1\right)$prove

Q15. Prove that ( cos $\theta +i\mathrm{sin}\theta $)

^{n}= cos $\left(\mathrm{n}\mathrm{\theta}\right)+i\mathrm{sin}(\mathrm{n}\mathrm{\theta})$, for all n $\in $ N by using PMI.$Provethat(1+x{)}^{n}\ge \left(1+nx\right),forallnaturalnumbern,wherex-1bymathematicalinduction.$

$\frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}{\displaystyle +}{\displaystyle \sqrt{a}}},\frac{1}{\sqrt{a}{\displaystyle +}{\displaystyle \sqrt{b}}}arein\phantom{\rule{0ex}{0ex}}\left(A\right)A.P.\left(B\right)G.P.\phantom{\rule{0ex}{0ex}}\left(C\right)H.P.\left(D\right)Noneofthese$

Please explain the last line of solution

2n?>?n2, where?n?is a positive integer such that?n?> 4.

Solution:

Let the given statement be P(n), i.e.,

P(n) : 2n?>?n2?where?n?> 4

For?n?= 5,

25?= 32 and 52?= 25

?25?> 52

Thus, P(n) is true for?n?= 5.

Let P(n) be true for?n?=?k, i.e.,

2k?>?k2?? (1)

Now, we have to prove that P(k? 1) is true whenever P(k) is true, i.e. we have to prove that 2k? 1?> (k? 1)2.

From equation (1), we obtain

2k?>?k2

On multiplying both sides with 2, we obtain

2 ? 2k?> 2 ??k2

2k? 1?> 2k2

?To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2.

Let us assume 2k2?> (k? 1)2.

? 2k2?>?k2? 2k? 1

??k2?> 2k? 1

??k2?? 2k?? 1 > 0

? (k?? 1)2?? 2 > 0

? (k?? 1)2?> 2, which is true as?k?> 4

Hence, our assumption 2k2?> (k? 1)2?is correct and we have 2k? 1?> (k? 1)2.

Thus, P(n) is true for?n?=?k? 1.

Thus, by the principle of mathematical induction, the given mathematical statement is true for every positive integer?n.

?

IN THIS EXAMPLE IT HAS BEEN WRITTEN THAT ,

To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2.

how?2k? 1?> (k? 1)2?=?2k2?> (k? 1)2

I cant understand how so plz explain that