prove by induction that ( 2n + 7 ) < (n + 3)^2 for all natural numbers n. using this prove by induction that  for all n=N  , sin theta + sin2 theta + sin 3 theta + ......sin n theta = sin( n+ 1 )/2 theta sin n theta/2 / sin theta / 2 ​

Dear student

Assuming that you  know how to prove P(1) is true, let us start with p(k)

(2k+7)<(k+3)^2           (Add 2 on both sides)  (Because we have to prove that p(k+1) is also true)
2k+7+2 < k^2 + 6k + 9 +2
2k+9 < k2 + 6k + 11 -------------     (4)

Know in rough subsitute the valu of p(n)= k+1
you will get = k2  + 8k + 16 in RHS 
And LHS would be 2k + 9

we know 8k > 6k  ------(1)
               16 > 11 -------(2)
Add (1) and (2)
                        8k + 16 > 6k + 11
Add k2 on both sides
                        k2 + 8k + 16 > k2 + 6k + 11 --------------(3)

From (3) and (4)                                                    
                       2k + 9 > k2 + 8k + 16
Which is equal to P(k + 1).................
Therefore it is true for all n E N
Regards

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