if (n=*+j*j+k*k*k) printf(%4d"n) } printf(n) 4求艺+2+头 maino int nl.n2.n3: nl-100,n2-50,n3=10 int k(floatk.) f1 oat sl=0,s2-=0,s3=0,s-0 for (k=1;k<=n1;k++) sl=s1+k: for(k=l:k<=n2.k++) s2=s2+k; for(k=l:k<=n3.k++) s3=s3+1/k,s-s1+s2+s3: printf(“s-%8.2n”,sh S=47977.93 5.用牛倾迭代法求下列方程在1.5附近的根 2x3.4x2+3x-6-0 #include <math.h main() (floatx.x0.f.fl x=1.5, % {x0=x f(2*x0-4)*x0+3)*x0-6 f1=6*x0-8)*x0+3 while (fabs(x-x0)>=(e-5); printf("The root of requation is%5.2fn").if (n==i*i*i+j*j*j+k*k*k) printf(“%4d”,n); } printf(“\n”); } 4.求 s= = 100 k 1 k + = 50 1 2 k k + = 10 1 1 k k 。 main() { int n1,n2,n3; n1=100;n2=50;n3=10; int k; (float k;) float s1=0,s2=0,s3=0,s=0; for (k=1;k<=n1;k++) s1=s1+k; for (k=1;k<=n2;k++) s2=s2+k; for (k=1;k<=n3;k++) s3=s3+1/k; s=s1+s2+s3; printf(“s=%8.2f\n”,s); } S=47977.93 5.用牛顿迭代法求下列方程在 1.5 附近的根。 2x 3 -4x 2 +3x-6=0 #include <math.h> main() { float x,x0,f,f1; x=1.5; do { x0=x; f=((2*x0-4)*x0+3)*x0-6 f1=(6*x0-8)*x0+3; } while (fabs(x-x0)>=(e-5); printf(“The root of requation is %5.2f\n”);