[tex]\sf \pink {Maths\: legends\: need \:help…!! }[/tex]
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[tex]\large \red ✿\mathtt{\bold{ Prove\: tha

2 thoughts on “[tex]\sf \pink {Maths\: legends\: need \:help…!! }[/tex]<br />[tex]\\\\[/tex]<br />[tex]\large \red ✿\mathtt{\bold{ Prove\: tha”

1. Step-by-step explanation:

   let p(n):(1+3/1)(1+5/4)(1+7/9)(1+2n+1/n^2)=(n+1)^2

   for n=1 :lhs=(1+3/1)=4.

   Rhs=(1+1)^2=4

   therefore,p(1)=true

   let us assume p(k) is true for some k is element of N

   i.e (1+3/1)(1+5/4)(1+7/9)(1+2k+1/n^2)=(k+1)^2

   =(k+1)^2[1+2k+3/(k+1)^2]=(k+1)^2 + 2k + 3

   =k^2+4k+4=[k+2]^2 = (k + 1) ^+ 2k + 3

   =k^2+4k+4=[k+2]^2= [(k + 1)+1]^2

   which is P(k+1)

   hence by mathematical induction p(n) is true for all

   n E N

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2. Solution :-

   [tex]\displaystyle\sf P(n): \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2n+1}{n^2}\right)=(n+1)^2[/tex]

   For n = 1,

   [tex] : \implies \sf P(1) = \left(1+\dfrac{(2 \times 1)+1}{1^2}\right)= (1+1)^2 [/tex]

   [tex]: \implies \sf P(1) = \left(1+\dfrac{3}{1}\right)= 2^2[/tex]

   [tex]: \implies \sf P(1) =4= 4[/tex]

   Thus P(n) is true for n = 1.

   Assume that P(k) is true.

   ie,

   [tex] \sf \displaystyle \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2k+1}{k^2}\right)=(k+1)^2[/tex]

   We shall now prove that P(k+1) is true.

   [tex] : \implies \sf LHS \: of \: P(k+1) =(k + 1) ^{2} \cdot \left( 1 + \dfrac{2k + 3}{ {(k + 1)}^{2} } \right)[/tex]

   [tex]: \implies \sf LHS \: of \: P(k+1) =(k + 1) ^{2} \cdot \left( \dfrac{(k + 1) ^{2} + 2k + 3}{(k + 1) ^{2} } \right)[/tex]

   [tex] : \implies \sf LHS \: of \: P(k+1) =(k + 1) ^{2} + 2k + 3[/tex]

   [tex]: \implies \sf LHS \: of \: P(k+1) = {k}^{2} + 2k + 1 + 2k + 3[/tex]

   [tex]: \implies \sf LHS \: of \: P(k+1) = {k}^{2} + 4k + 4[/tex]

   [tex] : \implies \sf LHS \: of \: P(k+1) =(k + 2) ^{2} [/tex]

   [tex]: \implies \sf LHS \: of \: P(k+1) =(\overline{k + 1 }+ 1) ^{2} [/tex]

   [tex]: \implies \sf LHS \: of \: P(k+1) =RHS \: of \: P(k+1)[/tex]

   Thus P(k+1) is true whenever P(k) is true.

   Hence, by the principle of mathematical induction P(n) is true for all n∈N.

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