20. 2, 4, & 16, 32, 64 21.4 7, 10, 13, 16 22 256, 128 64, 32,16. 23, For questio

Question

20. 2, 4, & 16, 32, 64 21.4 7, 10, 13, 16 22 256, 128 64, 32,16. 23, For questions No 20 and Na 21, () What are te nesct two lormWhat is the Councs Diffe ence er Cotnaos Rstio for each" 24 Uing Successive Dalerence, wait is the ned tem in each of the Solloving? ) 1, , 22, 37, 56, i 0, 12, 72, 240, 600, 1260, 2352 to find the firt o 25b. Find the eormon dittierence 23a Usethe foemul in 25a $-3-1-2 260. rind the T*term of L-2-5. 360) Fiad the 5 tenm of 7 Complete the following table Triangular Square Pentagonal Hexagonal B. Proof by Induction Prove the following by induction 1+3+5+7+ +(2n-1)=r BONUS See eample below for belp Exumple Proof by Induction

Explanation / Answer

(20)

the given series is

2,4,8,16,32,64,..., .....

the next 2 term of the series is

128 and 256

and common ratio is

4/2=8/4=16/8=....256/128=2

(21)

the given series is

1,4,7,10,13,16,...,....

the next 2 term of the series is

19 and 22

and the common diffirence is

4-1=7-4=10-7=....=22-19=3

(22)

256,128,64,32,16,...,....

the next 2 term of the series is

8 and 4

and the common ratio is

128/256=64/128=32/64...=1/2

(question)

example :

to proof using induction

1+3+5+7+......+(2n-1)=n2

proof:==>

For n=1 this is true, since 1 = 12

Suppose this is true for some n = k , that is

1+3+5+7+......+(2k-1)=k2

now prove that this is true for n = k + 1, that is

1+3+5+7+......+(2k-1)+(2k+1)=(k+1)2

now we have

1+3+5+7+......+(2k-1)+(2k+1)=k2+(2k+1)

1+3+5+7+......+(2k-1)+(2k+1)=(k+1)2

hence its prooved.....

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