for all natural numbers n ≥ 2.
given;let ⇒ p(k+1) is true when p(k) is true.∴ by mathematical induction is true for all natural number n≥2.
2, 0 ( any two integers with one of them as 0). answer: k=7 [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] 2 use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect ha = 2hl \[\ce{fecl2 +2naoh -> fe(oh)2 + 2naci}\] 1 a c d
2 a
3 c
4 b
5 c
i got a 100% tan a/2 is equal to if , passing through (1, 2), satisfies the differential equation , then