Question

induction problem

P + +8t...n = 12 (n +32 4 + 4 +7 + 3n-1 + (3 -2) = n @ren) = Že - (160)

prove it

Answer 1

Answer:--------------------

3.
proof that 1³+2³+3³+…+n³ = n²(n+1)² / 4
Let S = 1³+2³+3³+…+n³

In this proof, we assume the existence of the following, known results:

1+2+3+…+n=n(n+1) / 2.............................…(A)

1²+2²+3²+…+n²=n(n+1)(2n+1) / 6…............................(B)

Let us now start by writing the following identity.

n⁴−(n−1)⁴=4n³−6n²+4n−1…........................................(1)

Replacing n with (n−1), we obtain

(n−1)⁴−(n−2)⁴=4(n−1)³−6(n−1)²+4(n−1)−1......................…(2)

Replacing n with (n−2) in (1) above,

(n−2)⁴−(n−3)⁴=4(n−2)³−6(n−2)²+4(n−2)−1.....................................…(3)

……

Replacing n with n−(n−3), i.e. setting n=3 in (1) above,

3⁴−2⁴=4.3³−6.3²+4.3 –1…................(n−2)

Similarly, with n=2,(1) gives

2⁴−1⁴=4.2³−6.2²+4.2–1…...................................... (n−1)

And finally, when n=1, we have

1⁴−0⁴=4.1³−6.1²+4.1–1................................................ (n)

Adding all of these n equations starting from (1),(2),(3),…(n), we obtain

n⁴=4[n³+(n−1)³+(n−2)³+…+3³+2³+1³]−6[n²+(n−1)²+(n−2)²+…+3²+2²+1²]+4[n+(n−1)+(n−2)+…+3+2+1]−(1+1+1+…+1)

=4S− 6×n(n+1)(2n+1) / 6+ 4×n(n+1)/ 2−n

from (A) and (B)

=4S− n(n+1)(2n+1)+2n(n+1)−n
=4S−n²(2n+1)

∴4S=n⁴+2n³+n²=n²(n+1)²

from which the result follows:

S=n²(n+1)² / 4
1³+2³+3³+…+n³ = n²(n+1)² / 4
Hence proved.

4.
P(n) = 1 + 4 + 7+ ----- + (3n - 2) =n (3n-1) / 2
For n = 1,
P (1) = 1 (3-1) / 2 = 1 which is true.
Assume that P(k) is true for some positive integers k
P(k): 1 + 4 + 7 + ----- + (3k - 2) =k(3k-1) / 2
We shall prove that P(k + 1) is also true.
Now, we have
1 + 4 + 7 + ----- (3k - 2) + [3(k + 1) - 2] = k(3k-1) / 2 + [3(k + 1) - 2]
== 3k² - k / 2 + ( 3k+1 )
== ( 3k² - k + 6k + 2 ) / 2
== ( 3k² + 5k + 2 ) / 2
== ( 3k + 2 ) ( k + 1 ) / 2
Thus P(k +1) is true, whenever P(k) is true.
Hence from principle of mathematical induction, the statement P(n) is true for all natural numbers N.