1,Tính các tổng sau. a) 1 + 2+ 3+ 4 +....+ nb) 2+4+6+8+...+2.nc) 1+3+5+7+...+(2.n +1)d) 1+4+7+10+..+2005e) 2+5+8+...+2006f) 1+5+9+..+20012,Tính nhanh : A = 1 +2 + 4 + 8 +16 + ...+ 8192 3,a, Tính tổng các số lẻ có 2 chữ số.b,Tính tổng các số chẵn có 2 chữ số.4,a,Tổng 1 +2+3+....+n có bao nhiêu số hạng để kết quả tổng bằng 190b,Có hay không số tự nhiên n sao cho 1+2+3+...+n =2004c,Chứng minh rằng: [(1+2+3+...+n)-7]không chia hết cho 10

Câu 6:

uses crt;

var n,i:integer;

begin

clrscr;

readln(n);

for i:=1 to n do 

if n mod i=0 then write(i:4);

readln;

end.

5:

uses crt;

var n,i,dem:integer;

begin

clrscr;

readln(n);

dem:=0;

for i:=0 to n do

if i mod 2=1 then 

begin

write(i:4);

dem:=dem+1;

end;

writeln;

writeln(dem);

readln;

end.

câu 1

Câu hỏi của Ngọc Hà - Toán lớp 6 - Học toán với OnlineMath

a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)] 

=n.(n+1).(n+2) 

=>S=[n.(n+1).(n+2)] : 3

bb

Số số hạng của A là:100-1+1=100(số)

Tổng của A là:

(100+1).100:2=5050

Tổng quát: A=1+2+3+...+n=(n+1).n:2

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1)n[(n+1)-(n-2)]
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+(n-1)n(n+1)-(n-2)(n-1)n
3A=(n-1)n(n+1)
A=(n-1)n(n+1)/3

Ta có : 

\(A=1.2+2.3+3.4+...+\left(n-1\right).n\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+\left(n-1\right).n.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right).n.\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+\left(n-1\right)n.\left(n+1\right)-\left(n-2\right).\left(n-1\right).n\)

\(\Rightarrow3A=\left(n-1\right).n.\left(n+1\right)\)

\(\Rightarrow A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)

Vậy \(A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)

P/s : Mik ko chắc 

~ Ủng hộ nhé 

ta có: \(A=1.2+2.3+3.4+...+\left(n-1\right).n\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+\left(n-1\right).n.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right).n.\left[\left(n+1\right)-[\left(n-1\right)-1]\right]\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.34+...+\left(n-1\right).n.\left(n+1\right)-\left[\left(n-1\right)-1\right].\left(n-1\right).n\)

\(3A=\left(n-1\right).n.\left(n+1\right)\)

\(A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)

int Sum(int n){

int temp=1;

int result=0;

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

{

result+=i*temp;

temp*=-1;

}

return result;

}

 

#include <bits/stdc++.h>using namespace std;const int maxn = 1e2 + 14, lg = 15; /*  ######################################################################  #######################   THE   BIG   INT   ##########################*/const int base = 1000000000;const int base_digits = 9;struct bigint {vector<int> a;int sign;/*<arpa>*/int size(){if(a.empty())return 0;int ans=(a.size()-1)*base_digits;int ca=a.back();while(ca)ans++,ca/=10;return ans;}bigint operator ^(const bigint &v){bigint ans=1,a=*this,b=v;while(!b.isZero()){if(b%2)ans*=a;a*=a,b/=2;}return ans;}string to_string(){stringstream ss;ss << *this;string s;ss >> s;return s;}int sumof(){string s = to_string();int ans = 0;for(auto c : s)  ans += c - '0'return ans;}/*</arpa>*/bigint() :sign(1) {} bigint(long long v) {*this = v;} bigint(const string &s) {read(s);} void operator=(const bigint &v) {sign = v.sign;a = v.a;} void operator=(long long v) {sign = 1;a.clear();if (v < 0)sign = -1, v = -v;for (; v > 0; v = v / base)a.push_back(v % base);} bigint operator+(const bigint &v) const {if (sign == v.sign) {bigint res = v; for (int i = 0, carry = 0; i < (int) max(a.size(), v.a.size()) || carry; ++i) {if (i == (int) res.a.size())res.a.push_back(0);res.a[i] += carry + (i < (int) a.size() ? a[i] : 0);carry = res.a[i] >= base;if (carry)res.a[i] -= base;}return res;}return *this - (-v);} bigint operator-(const bigint &v) const {if (sign == v.sign) {if (abs() >= v.abs()) {bigint res = *this;for (int i = 0, carry = 0; i < (int) v.a.size() || carry; ++i) {res.a[i] -= carry + (i < (int) v.a.size() ? v.a[i] : 0);carry = res.a[i] < 0;if (carry)res.a[i] += base;}res.trim();return res;}return -(v - *this);}return *this + (-v);} void operator*=(int v) {if (v < 0)sign = -sign, v = -v;for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {if (i == (int) a.size())a.push_back(0);long long cur = a[i] * (long long) v + carry;carry = (int) (cur / base);a[i] = (int) (cur % base);//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));}trim();} bigint operator*(int v) const {bigint res = *this;res *= v;return res;} void operator*=(long long v) {if (v < 0)sign = -sign, v = -v;if(v > base){*this = *this * (v / base) * base + *this * (v % base);return ;}for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {if (i == (int) a.size())a.push_back(0);long long cur = a[i] * (long long) v + carry;carry = (int) (cur / base);a[i] = (int) (cur % base);//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));}trim();} bigint operator*(long long v) const {bigint res = *this;res *= v;return res;} friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1) {int norm = base / (b1.a.back() + 1);bigint a = a1.abs() * norm;bigint b = b1.abs() * norm;bigint q, r;q.a.resize(a.a.size()); for (int i = a.a.size() - 1; i >= 0; i--) {r *= base;r += a.a[i];int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];int d = ((long long) base * s1 + s2) / b.a.back();r -= b * d;while (r < 0)r += b, --d;q.a[i] = d;} q.sign = a1.sign * b1.sign;r.sign = a1.sign;q.trim();r.trim();return make_pair(q, r / norm);} bigint operator/(const bigint &v) const {return divmod(*this, v).first;} bigint operator%(const bigint &v) const {return divmod(*this, v).second;} void operator/=(int v) {if (v < 0)sign = -sign, v = -v;for (int i = (int) a.size() - 1, rem = 0; i >= 0; --i) {long long cur = a[i] + rem * (long long) base;a[i] = (int) (cur / v);rem = (int) (cur % v);}trim();} bigint operator/(int v) const {bigint res = *this;res /= v;return res;} int operator%(int v) const {if (v < 0)v = -v;int m = 0;for (int i = a.size() - 1; i >= 0; --i)m = (a[i] + m * (long long) base) % v;return m * sign;} void operator+=(const bigint &v) {*this = *this + v;}void operator-=(const bigint &v) {*this = *this - v;}void operator*=(const bigint &v) {*this = *this * v;}void operator/=(const bigint &v) {*this = *this / v;} bool operator<(const bigint &v) const {if (sign != v.sign)return sign < v.sign;if (a.size() != v.a.size())return a.size() * sign < v.a.size() * v.sign;for (int i = a.size() - 1; i >= 0; i--)if (a[i] != v.a[i])return a[i] * sign < v.a[i] * sign;return false;} bool operator>(const bigint &v) const {return v < *this;}bool operator<=(const bigint &v) const {return !(v < *this);}bool operator>=(const bigint &v) const {return !(*this < v);}bool operator==(const bigint &v) const {return !(*this < v) && !(v < *this);}bool operator!=(const bigint &v) const {return *this < v || v < *this;} void trim() {while (!a.empty() && !a.back())a.pop_back();if (a.empty())sign = 1;} bool isZero() const {return a.empty() || (a.size() == 1 && !a[0]);} bigint operator-() const {bigint res = *this;res.sign = -sign;return res;} bigint abs() const {bigint res = *this;res.sign *= res.sign;return res;} long long longValue() const {long long res = 0;for (int i = a.size() - 1; i >= 0; i--)res = res * base + a[i];return res * sign;} friend bigint gcd(const bigint &a, const bigint &b) {return b.isZero() ? a : gcd(b, a % b);}friend bigint lcm(const bigint &a, const bigint &b) {return a / gcd(a, b) * b;} void read(const string &s) {sign = 1;a.clear();int pos = 0;while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {if (s[pos] == '-')sign = -sign;++pos;}for (int i = s.size() - 1; i >= pos; i -= base_digits) {int x = 0;for (int j = max(pos, i - base_digits + 1); j <= i; j++)x = x * 10 + s[j] - '0'a.push_back(x);}trim();} friend istream& operator>>(istream &stream, bigint &v) {string s;stream >> s;v.read(s);return stream;} friend ostream& operator<<(ostream &stream, const bigint &v) {if (v.sign == -1)stream << '-'stream << (v.a.empty() ? 0 : v.a.back());for (int i = (int) v.a.size() - 2; i >= 0; --i)stream << setw(base_digits) << setfill('0') << v.a[i];return stream;} static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {vector<long long> p(max(old_digits, new_digits) + 1);p[0] = 1;for (int i = 1; i < (int) p.size(); i++)p[i] = p[i - 1] * 10;vector<int> res;long long cur = 0;int cur_digits = 0;for (int i = 0; i < (int) a.size(); i++) {cur += a[i] * p[cur_digits];cur_digits += old_digits;while (cur_digits >= new_digits) {res.push_back(int(cur % p[new_digits]));cur /= p[new_digits];cur_digits -= new_digits;}}res.push_back((int) cur);while (!res.empty() && !res.back())res.pop_back();return res;} typedef vector<long long> vll; static vll karatsubaMultiply(const vll &a, const vll &b) {int n = a.size();vll res(n + n);if (n <= 32) {for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)res[i + j] += a[i] * b[j];return res;} int k = n >> 1;vll a1(a.begin(), a.begin() + k);vll a2(a.begin() + k, a.end());vll b1(b.begin(), b.begin() + k);vll b2(b.begin() + k, b.end()); vll a1b1 = karatsubaMultiply(a1, b1);vll a2b2 = karatsubaMultiply(a2, b2); for (int i = 0; i < k; i++)a2[i] += a1[i];for (int i = 0; i < k; i++)b2[i] += b1[i]; vll r = karatsubaMultiply(a2, b2);for (int i = 0; i < (int) a1b1.size(); i++)r[i] -= a1b1[i];for (int i = 0; i < (int) a2b2.size(); i++)r[i] -= a2b2[i]; for (int i = 0; i < (int) r.size(); i++)res[i + k] += r[i];for (int i = 0; i < (int) a1b1.size(); i++)res[i] += a1b1[i];for (int i = 0; i < (int) a2b2.size(); i++)res[i + n] += a2b2[i];return res;} bigint operator*(const bigint &v) const {vector<int> a6 = convert_base(this->a, base_digits, 6);vector<int> b6 = convert_base(v.a, base_digits, 6);vll a(a6.begin(), a6.end());vll b(b6.begin(), b6.end());while (a.size() < b.size())a.push_back(0);while (b.size() < a.size())b.push_back(0);while (a.size() & (a.size() - 1))a.push_back(0), b.push_back(0);vll c = karatsubaMultiply(a, b);bigint res;res.sign = sign * v.sign;for (int i = 0, carry = 0; i < (int) c.size(); i++) {long long cur = c[i] + carry;res.a.push_back((int) (cur % 1000000));carry = (int) (cur / 1000000);}res.a = convert_base(res.a, 6, base_digits);res.trim();return res;}};/*  #######################   THE   BIG   INT   ##########################  ######################################################################*/bigint n;int main(){ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0);cin>>n;cout<<"-"<<(n+1)/2;return 0;}

Vì ko biết Giới hạn đến bao nhiêu nên mình làm thế này é!