Let the given statement be P(n). Then,
`P(n): (ab)^(n) = a^(n) b^(n)`.
When n = 1 , we have
LHS = `(ab)^(1) = ab and RHS = a^(1)b^(1) = ab`.
`:. ` LHS = RHS.
Thus , the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
`P(k) : (ab)^(k) = a^(k) b^(k)`. ….(i)
Now, `(ab)^(k+1) = (ab) ^(k) (ab) = (a^(k)b^(k))(ab)` [using (i)]
` = (a^(k)*a)(b^(k)*b)` [by commnutativity and associativity of multiplication on real numbers]
`(a^(k+1)*b^(k+1))`.
`:. ” ” P(k+1): (ab)^(k+1)=(a^(k+1)*b^(k+1))`.
This shows that P(k+1) is true , whenever P(k) is true.
` :. ` P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
`(ab)^(n) = a^(n) b^(n) ” for all ” x in N`.

Let
`P (n) : (1)/(2.5)+(1)/(5.8)+(1)/(8.11)+…..+(1)/((3n-1)(3n+2))`
`=(n)/((6n+4))`
for n=1
`L.H.S. =(1)/(2.5)+(1)/(10)`
and `R.H.S. =(1)/(6.1+4) =(1)/(6+4)=(1)/(10)`
`rArr ” “L.H.S. =R.H.S.`
Therefore given statement is true for n=1
Let the statement P (n) be true for n=k
`:. P(k) =(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+……`
`+(1)/((3k-1)(3k+2))=(k)/(6k+4)`
For n=K+1
`P(K+1) : (1)/(2.5)+(1)/(5,8)+(1)/(8.11) +…….`
`+(1)/[[(3k+1)-1][3(k+1)+2)]`
`=(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+…..+(1)/((3k-1)(3k+2))`
`+(1)/((3k+2)(3k+5))`
`((1)/(2.5)+(1)/(5.8)+(1)/(8.11)+………+(1)/((3k-1)(3k+2)))`
`+(1)/((3k+2)(3k+5))`
`=(k)/(6k+4)+(1)/((3k+2)(3k+5))`
`=(1)/((3k+2))((K)/(2)+(1)/(3k+5))`
`=(1)/(3k+2)[(3k^(2)+5k+2)/(6k+10)]`
`=(1)/(3k+2)[(3k^(2)+3k+2K+2)/(6k+10]]`
`=(1)/(3k+2).[(3k(K+1)+2(k+1))/(6k+10)]`
`=(1)/(3k+2).((3k+2)(k+1))/(6k+10)=(k+1)/(6k+10)`
Then given statment P (n) is also true for n=K+1
Hence given statement P (n) is true for all values of n where `n in N`

Let the given statement be P(n). Then,
`P(n): 1*3+3*5+5*7+…+(2n-1)(2n+1)= 1/3 n(4n^(2)+6n-1)`.
When n = 1, we have
LHS = ` 1*3 and RHS = 1/3 xx 1 xx(4xx1^(2)+6xx1-1)=1/3 xx 1 xx 9 = 3`.
` :. ` LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
`P(k): 1*3+3*5+5*7+…+(2K-1)(2k+1)= 1/3 k(4k^(2)+6k-1).” “` …(i)
Now, ` 1*3+3*5+5*7+…+(2k-1)(2k+1)+{2(k+1)-1}{2(k+1)+1}`
` ={1*3+3*5+5*7+…+(2k-1)(2k+1)}+(2k+1)(2k+3)`
` = 1/3 k(4k^(2)+6k-1)+(2k+1)(2k+3)” “`[using (i)]
` = 1/3 [(4k^(3)+6k^(2)-k)+3(4k^(2)+8k+3)] = 1/3 (4k^(3)+18k^(2)+23k+9)`
` = 1/3 (k+1)(4k^(2)+14k+9)= 1/3 (k+1)[4(k+1)^(2)+6(k+1)-1]`.
` :. P(k+1): 1*3+3*5+5*7+…+{2(k+1)-1}{2(k+1)+1}`
` = 1/3 (k+1){4(k+1)^(2)+6(k+1)-1}`.
Thus, P(k+1) is true, whenever P(k) is true.
`:. ` P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction , we have
` 1*3+3*5+5*7+…+(2n-1)(2n+1) = 1/3 n (4n^(2)+6n-1)” for all ” n in N`.

Let P(n) : `n^(3)-7n+3` is divisible by 3, for all natural number n.
Step I We observe that P(1) is true.
`P(1)=(1)^(3)-7(1)+3`
=11-7+3
-3, which is divisible by 3.
Hence, P(1) is true. Step II Now, assume that P(n) is true for n=k.
`P(k+1):(k+1)^(3)-7(k+1)+3`
`=k^(3)+1+3k(k+1)-7k-7+3`
`=k^(3)-7k+3+3k(k+1)-6`
`=3q+3[k(k+1)-2]`
Hence, P(k+1) is true whenever P(k) is true. [from stem II]
So, by the principle of mathematical induction P(n): is true for all natural number n.

Let the given statement be P(n). Then,
` P(n) : 1/(3*5)+1/(5*7) + 1/(7*9) +…+ 1/((2n+1)(2n+3)) = n/(3(2n+3))`.
Putting n = 1 in the given statement, we get
LHS ` = 1/(3*5) = 1/15 and RHS = 1/(3(2xx1+3))= 1/15`.
` :. ` LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
`P(k): 1/(3*5)+1/(5*7)+1/(7*9) + …+1/((2k+1)(2k+3)) = k/(3(2k+3)).`…(i)
Now, ` 1/(3*5) + 1/(5*7) + …+1/((2k+1)(2k+3))+1/({2(k+1)+1}{2(k+1)+3}) `
`{1/(3*5)+1/(5*7)+..+1/((2k+1)(2k+3))}+1/((2k+3)(2k+5))`
` = k/(3(2k+3))+1/((2k+3)(2k+5)) ” “` [using (i)]
`=(k(2k=5)+3)/(3(2k+3)(2k+5))= ((2k^(2)+5k+3))/(3(2k+3)(2k+5))=((k+1)(2k+3))/(3(2k+3)(2k+5)) `
`=((k+1))/(3(2k+5))= ((k+1))/(3{2(k+1)+3}).`
`:. ” ” P(k+1) : 1/(3*5)+1/(5*7)+1/(7*9)+…+1/({2(k+1)+1}{2(k+1)+3}) = ((k+1))/(2{2(k+1)+3}).`
This shows that P(k+1) is true, whenever P(k) is true.
`:. ` P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
` 1/(3*5)+1/(5*7)+1/(7*9)+…+ 1/((2n+1)(2n+3)) = n/(3(2n+3)) ” for all values of ” n in N`.