Question

$\sf \pink {Maths\: legends\: need \:help...!! }$

$\large \red ✿\mathtt{\bold{ Prove\: that:-}}$

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

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Answer 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

Answer 2

Solution :-

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

For n = 1,

$: \implies \sf P(1) = \left(1+\dfrac{(2 \times 1)+1}{1^2}\right)= (1+1)^2$

$: \implies \sf P(1) = \left(1+\dfrac{3}{1}\right)= 2^2$

$: \implies \sf P(1) =4= 4$

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

Assume that P(k) is true.

ie,

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

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

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

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

$: \implies \sf LHS \: of \: P(k+1) =(k + 1) ^{2} + 2k + 3$

$: \implies \sf LHS \: of \: P(k+1) = {k}^{2} + 2k + 1 + 2k + 3$

$: \implies \sf LHS \: of \: P(k+1) = {k}^{2} + 4k + 4$

$: \implies \sf LHS \: of \: P(k+1) =(k + 2) ^{2}$

$: \implies \sf LHS \: of \: P(k+1) =(\overline{k + 1 }+ 1) ^{2}$

$: \implies \sf LHS \: of \: P(k+1) =RHS \: of \: P(k+1)$

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.