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author | smlckz <smlckz@college> | 2021-12-22 14:56:13 +0530 |
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committer | smlckz <smlckz@college> | 2021-12-22 14:56:13 +0530 |
commit | b73983c3717642ca10e7cfe93d97609adc377da9 (patch) | |
tree | a6e9fe4c27e3caa215f8aefa9265fb52f6de4375 /assignments/19-int-series.c | |
download | college-b73983c3717642ca10e7cfe93d97609adc377da9.tar.gz |
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Diffstat (limited to 'assignments/19-int-series.c')
-rw-r--r-- | assignments/19-int-series.c | 46 |
1 files changed, 46 insertions, 0 deletions
diff --git a/assignments/19-int-series.c b/assignments/19-int-series.c new file mode 100644 index 0000000..7403517 --- /dev/null +++ b/assignments/19-int-series.c @@ -0,0 +1,46 @@ +#include <stdio.h> + +int main(void) +{ + int i, sum, n, a; + printf("Calculate the sum of (-1)^(n+1) * i for i from 1 to given number.\n\n"); + printf("Enter a number: "); + scanf("%d", &n); + a = 1; + sum = 0; + for (i = 1; i <= n; i++) { + sum += a * i; + a = -a; + } + printf("The sum is %d\n", sum); + return 0; +} + +/* +Output: +Set 1: +Calculate the sum of (-1)^(n+1) * i for i from 1 to given number. + +Enter a number: 1 +The sum is 1 + +Set 2: +Calculate the sum of (-1)^(n+1) * i for i from 1 to given number. + +Enter a number: 2 +The sum is -1 + +Set 3: +Calculate the sum of (-1)^(n+1) * i for i from 1 to given number. + +Enter a number: 10 +The sum is -5 + +Set 4: +Calculate the sum of (-1)^(n+1) * i for i from 1 to given number. + +Enter a number: 123 +The sum is 62 + +*/ + |