Fill in the blanks by mathematical induction
1, prove by mathematical induction that "(3n+ 1) 7N- 1 is divisible by 9 (n? The second step of n) should be _ _ _ _ _.
2. Prove the equation "1+2+3+…+(n+3) = (nn)" by mathematical induction.
When n= 1, the left should be _ _ _ _ _ _ _.
3. Given that the first n terms of {an} sequence Sn=2n-an, then the first four terms of {an} are _ _ _ _ _ _ _ _ _ _ _.
4. When proving a proposition by mathematical induction, the formula on the left is (n is a positive even number) from "n=2k to n=2k+2", and the algebraic formula to be added on the left is _ _ _ _ _.
5. When using mathematical induction to prove that1+2+3+…+(2n+1) = (n+1), from "n=k to n=k+ 1".
6. Prove that 1+2+3+…+n = (n? The second step of n) should be; Assuming that the equation holds when _ _ _ _ _ _ _ _ _ _ _, then when _ _ _ _ _ _, the left side =1+2+…+_ _ _ _ _ = (65438+.
7. Given the series {an}, a is constant and an=, Sn=a 1+a2+…+an, then S 1, S2 S3 S3 and S3 are all _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
8. When proving the equation by mathematical induction, when n= 1, the left term is; The item to be added from'' is.
9. When it is proved by mathematical induction that it is a multiple of 3 1 at that time, when n= 1, the original formula is, and the term to be added from time to time is.
10、
Prove by mathematical induction "when n? 2 and n? When n, the first step of "xn-nan-1x+(n-1) an is divisible by (x-a)2" should be _ _ _ _ _ _ _ _ _ _.
1 1, the known sequence {an} satisfies a 1=2a, and an=2a-(n? 2) The first step to prove an=a by mathematical induction is _ _ _ _ _ _ _ _ _ _.
12. Prove the equation1.3.5+3.5.7++(2n-1) (2n+1) (2n+3) = n (n+2) (2n2) by mathematical induction.
13. In the series {an}, Sn is the sum of its first n terms, and Sn=2an-2, then the four terms of this series are _ _ _ _ _ _ _ _. Guess Ann's formula is _ _ _ _.
14. The first step to prove that "55n+ 1+45n+2+35n is divisible by 1 1" should be written as follows: when n = _ _ _ _ _, 55n+/kloc.
15. Prove 1+3+6+...+= (n? The first step of n) should be: when n = _ _ _ _ _, left = _ _ _ _ _, right = _ _ _ _, ∴ left _ _ _ _ _.
16. In the second step of proving that "56n+5+76n+7 is divisible by 9" by mathematical induction, 56 (k+1)+5+76 (k+1)+7 should be transformed into _ _
17. Let the sum of the internal angles of the convex K polygon be f(k), then the sum of the internal angles of the convex k+ 1 polygon is f (k+1) = f (k)+_ _ _.
18, given the sequence {an} and a 1=, then a2, a3, a4 and A5A5A5 are _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
19, explore the expression A=(n- 1)n- 1)! +(n-2)(n-2)！ +…+2 2! + 1 1! (n> 1 and n∈N), when the first step is n = _ _ _ _ _ _ _ _ a = _ _ _ _ _ _.
20. when proving a proposition by mathematical induction, the formula on the left is1.2.3.4+2.3.4.5+n (n+1) (n+2) (n+3), from "n=k to n = k+60.
2 1, when proving a proposition by mathematical induction, if the left side of the proposition is1+++...+(n? N), then when n=k+ 1, the left side should be the left side when n=k plus _ _ _ _ _ _ _ _.
2. Prove that1+2+22+23+...+25n-1(n? N) is a multiple of 3 1, from "n=k? N = k+ 1 "The item to be added is _ _ _ _ _ _ _ _.
23. Let sk =, then sk+1= sk+_ _ _
24. Remember that every two sides of a plane intersect at two points, and any three parabolas without * * * points divide the plane into z regions, and the number of regions is f(n), then f (k+1) = f (k)+_ _.
25. There are k points on the straight line L (k? 2) The number of line segments determined by k points is marked as f(k), so after adding a point on L, the number of line segments will increase by _ _ _ _ _.
26. There were originally k circles on the plane, and their intersection number was f(k), so after adding the circle of k+ 1, the intersection number would increase by _ _ _ _ _ _ _.
27, the original k circles on the plane, the arc formed by their intersection * * * has f(k) segment, then increase the k+ 1 circle with two intersections with the first k circles, but not the intersection of the first k circles, then increase the arc of the first k circles by _ _ _ _ _ _ _.
28. There are k planes passing through a point, and any three or more planes have no straight line. These k planes divide the space into f(k), so k+ 1 plane divides the space into f (k+1) = f (k)+_ _ _.
29. There are k straight lines in the original plane. No two of these k straight lines are parallel to each other, and no three are at the same point. They are divided into f(k) line segments or rays, and then a straight line is added, and the divided line segments or rays increase by _ _ _ _ _ _ _ _.
30. If any three intersecting k straight lines on a plane divide the plane into f(k) parts, then k+ 1 straight lines divide the plane into f (k+1) = f (k)+_ _ _ _.
3 1, given that the sum of internal angles of convex K- polygon is f(k), the relationship between the sum of internal angles of convex K+ 1 and f(k) is f (k+1) = _ _ _ _ _ _.
32. Let the sequence {an} satisfy a 1=2 and an+ 1 = 2an+2. In the second step of proving an = 4.2n- 1-2 by mathematical induction, let n=k and the conclusion holds, that is, AK = 4.2k-65438.
Fill in the blanks with mathematical induction [answers]
1, answer: omitted.
2、 1+2+3+4
3、 1,
4、
5 、( 2k+2)(2k+3)
6. Answer: Omit.
7、
8、 1+2+3; (2k+2)+(2k+3)
9、 1+2+22+23+24; 25k+25k+ 1+25k+2+25k+3+25k+4。
10, when n=2, xn-nan-1x+(n-1) an = x2-2ax+a2 = (x-a)2 can be divisible by (x-a) 2.
1 1、a2=2a-=2a-=a=
12、 1 3 5= 15; 1 3 (2+4- 1)= 15
13、 2,4,8, 16; 2n
14、 0,5 1+42+30,22
15, 1, 1, 1, =, hold.
16、76(56k+5+76k+7)+(56-76) 56k+5
17、 π
18、
19、2, 1
20 、( k+ 1)(k+2)(k+3)(k+4)
2 1、 +++…+
22、25k+25k+ 1+…+25k+4
23、
24、2k+ 1
25、k
26、2k
27、2k
28、2k
29、2k+ 1
30、k+ 1
3 1、f(k)+
32、AK+ 1 = 2ak+2 = 2(4 2k- 1-2)+2 = 4 2k-2 = 4 2(k+ 1)- 1-2
Example 1 proves that the polynomial xn+1+(x+1) 2n-1(n ∈ n) is divisible by the polynomial x2+x+ 1.
Analysis: propositions related to natural numbers are often proved by mathematical induction, but they are in use.
When proving divisibility by mathematical induction, it is often necessary to add and divide terms under the condition of n=k+ 1 in order to make assumptions.
It is proved that (1) when n= 1, x2+(x+ 1) is divisible by x2+x+ 1.
Example 2 proved by mathematical induction:
Comments: Usually, when proving a proposition with a natural number of n by mathematical induction, the first step is only to test n= 1 (or n=2, …). In this problem, the inequality of n= 1 is tested first, and then the inequality of n=2 is also established, which is not redundant, because it will be used in the later proof.
Example 3 It is known that n planes all pass through the same point, but none of them pass through the same straight line. It is proved that these n planes divide the space into f(n)=n(n- 1)+2 parts.
It is proved that (1) when n= 1, 1 plane divides the space into two parts, and f (1) =1× (1-kloc-0/)+2 = 2.
(2) Suppose that when n=k, the proposition holds, that is, k qualified planes divide the space into f(k)=k(k- 1)+2 (part).
When n=k+ 1, the plane k+ 1 intersects every other plane, which increases the space divided by it by two parts, so * * * increases the 2k part.
∴f(k+ 1)=f(k)+2k=k(k- 1)+2+2k
= k (k-1+2)+2 = (k+1) [(k+1)-1]+2 (partial),
That is, when n=k+ 1, the proposition holds.
According to (1) and (2), n qualified planes divide the space into f(n)=n(n- 1)+2 parts.