1/

\(N=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)=\)

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

\(\Rightarrow S=A-2B\)

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

\(N=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe

Câu 1 có sai đề bài không đấy?

Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)

                   \(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)

                       \(=6^2.1771=36.1771=63756\)

A = 1 +  (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)+...+ (1 + 2 + 3+...+ 30)

A = 1 + (2+1)\(\times\)2: 2+ (3+1)\(\times\)3: 2 + ....+ (30 +1) \(\times\) 30: 2 

A = 1 + 2 \(\times\)3: 2 + 3 \(\times\) 4 : 2 +...+ 30 \(\times\) 31:2

A = \(\dfrac{2+2\times3+3\times4+...+30\times31}{2}\)

Đặt B  = 2 + 2 \(\times\) 3 + 3 \(\times\) 4 +...+ 30 \(\times\) 31

   1 x 2 x 3 = 1 x 2 x3           = 1 x 2 x 3

   2 x 3 x 3 = 2 x 3 x ( 4 - 1) = 2 x 3 x 4 - 1 x 2 x 3

   3 x 4 x 3 = 3 x 4 x (5- 2)   = 3 x 4 x 5 - 2 x 3 x 4 

   ...........................................................................

   30 x 31 x 3 = 30 x 31 x (32 - 29) = 30 x 31 x 32 - 29 x 31 x 30

   Cộng vế với vế ta có: 

B x 3 = 30 x 31 x 32 ⇒ B = 30 x 31 x 32 : 3 = 14880

A = B : 2 = 14880 : 2 = 7440

\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ S=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)