**Click on the series to view the code**

1) 1 + 2 + 3 + 4 + 5 + ... + n

2) (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + ... + (n*n)

3) (1) + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+n)

4) 1! + 2! + 3! + 4! + 5! + ... + n!

5) (1^1) + (2^2) + (3^3) + (4^4) + (5^5) + ... + (n^n)

6) (1!/1) + (2!/2) + (3!/3) + (4!/4) + (5!/5) + ... + (n!/n)

7) [(1^1)/1] + [(2^2)/2] + [(3^3)/3] + [(4^4)/4] + [(5^5)/5] + ... + [(n^n)/n]

8) [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!]

9) 1/2 - 2/3 + 3/4 - 4/5 + 5/6 - ...... n

10) 1 2 3 6 9 18 27 54...

11) 2 15 41 80 132 197 275 366 470 587

12) 1 3 8 15 27 50 92 169 311

**Related Links:**

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2/9 - 5/13 + 8/17 - 11/21 + 14/25..... upto n terms

ReplyDeletefor(i=1;i<=n;i++)

Delete{

sum+=((i*3)-1)/((4*i)+5);

}

very nice

DeletePlz.... help to solve this problem

ReplyDeleteFind the sum of the series 5^2+ 〖10〗^2+ 〖15〗^(2 )+⋯

plz. solve it.

ReplyDelete1+3+5+7+.....+N

sum=0;

Deletefor(i=1;i<=n;i+=2)

sum+=i;

area of cicle ,reactangle .cylinder using swich statments

ReplyDeleteto generate the first n terms in the series --- 6,11,21,36,56,...

ReplyDeletea = 1;

Deleteb = 5

for (i=1; i<=n; i++)

{

printf("%d ",a+b);

b = b+5;

}

5 cannot be assumed it must got as input sir and btw wrong answer

Deletethen take input of x, start b = x and the b = b + x

Delete@Nimish Garg

DeleteThis program gives the sequence of 6,11,16,21,26 not 6,11,21,36,56

For the series 1,2,3,6,9,18,27....

ReplyDeletethank u in advanc

http://cbasicprogram.blogspot.in/2013/01/series-program-10.html

DeleteWrite a program to generate the first n terms in the series --- 2,15,41,80,...

ReplyDeletethank you in advance..

int a=2,i,n=10;

Deletefor(i=1;i<=n;i++)

{

printf("%d ",a);

a+=13*i;

}

http://cbasicprogram.blogspot.com/2013/01/series-program-11.html

Deleteplz give me the code for 1,3,4,8,15,27

ReplyDelete1+(1+3)+(1+3+5)....

ReplyDeletehelp me with this sequence: 110,24,1,3,9,2

ReplyDelete6 11 21 36 56 81

ReplyDeletea=6;

Deletefor(i=0;i<=5;i++)

{

a+=i*5;

printf("%d ",a);

}

sir, give me a general form.

DeleteFind sum of 1 + x + x2 + ... + xn without using pow()

ReplyDeletecalculate the sum of the n terms of the series s=1/2!+2/3!+3/4!+4/5!...

ReplyDeletei am also want this series,,,, if u have the program of this series plz sand meee....

Deleteplz help me the number sequence 120,99,80,65......

ReplyDeletecan anyone find sum of this series [x/1!+ x^2/2! +x^3/3! + x^4/4!+.......x^n/n!]

ReplyDeleteIN O(N) complexity

1*2+2*3+3*4+.....n*(n+1) help plss

ReplyDeletefor(i=1;i<n;i++)

Delete{

ans=i*(i+1);

printf("%d",ans);

}

please sir solve this questions

ReplyDeleteWrite a program in 'C' to compute the series :

(a) (x) + (x + n) + (x + n2) + (x + n3) + for a total of m terms. Where m, n and x are to

be accepted by the user.

(b) 1+5+11+17........n

S=1+a^2/3+a^4/15+a^6/35..a^n

ReplyDelete