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Ta có : k(k+1)(k+2)-(k-1)(k+1)k
=k(k+1).[(k+2)-(k-1)]
=3k(k+1)
áp dụng 3(1+2)=1.2.3-0.1.2
=>3(2.3)=2.3.4-1.2.3
=>3(3.4)=3.4.5-2.3.4
.....................................
3n(n+1)=n(n+1)(n+2)-(n-1)n(n+1)
Cộng lại ta có 3.S=n(n+1)(n+2)=>S=n(n+1)(n+2)/3
CHÚC BẠN HỌC TỐT NHA !!!
k(k+1)(k+2)-(k-1)k(k+1)=k(k+1)(k+2-k+1)=3.k.(k+1)
S=1.2+2.3+3.4+...+n(n+1)
=>3S=1.2.3+2.3.3+3.4.3+...+n(n+1)3
=1.2.3+2.3.(4-1)+3.4(5-2)+...+n.(n+1)[(n+2)-(n-1)]
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)
=n(n+1)(n+2)
\(\Rightarrow S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
m tưởng tao thik đăng à..............................................
k . (k+ 1) . (k+2) - k .(k +1) . (k-1)
= [ (k+2)-(k -1) ] .k .(k+1)
= (k + 2 -k +1) . k .(k+1)
= 3k (k+1)
Vậy: k . (k+ 1) . (k+2) - k .(k +1) . (k-1) = 3k (k+1)
S = 1.2+2.3+...+n.(n+1)
3S = 3.1.2 +3.2.3+...+3.n. (n+1)
3S = 1.2.3 - 0.1.2 +2.3.4 -1.2.3 + ... + n . (n+1 ) . (n+2) - (n-1).n.(n+1)
3S = n.(n+1).(n+2)
Ta có: k(k+1)(k+2)-(k-1)k(k+1)
=k(k+1)[(k+2)-(k-1)]
=k(k+1)[k+2-k+1]
=k(k+1)[(k-k)+(2+1)]
=k(k+1)3
=3k(k+1)
Vậy k(k+1)(k+2)-(k-1)k(k+1)=3k(k+1)
Áp dụng:
S=1.2+2.3+3.4+...+n(n+1)
3S=3.1.2+3.2.3+3.3.4+...+3.n(n+1)
3S=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)
3S=(1.2.3-1.2.3)+(2.3.4-2.3.4)+(3.4.5-3.4.5)+...+[(n-1)n(n+1)-(n-1)n(n+1)]+n(n+1)(n+2)-0
3S=n(n+1)(n+2)
S=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=k\left(k+1\right)\left(k+2-k+1\right)=3\)\(\)\(k\left(k+1\right)\left(DPCM\right)\)
\(S=1.2+2.3+3.4+....+n\left(n+1\right)\)
\(3S=3\left[1.2+2.3+...+n\left(n+1\right)\right]\)
\(3S=1.2.3-0.1.2+2.3.4-1.2.3+....+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(3S=n\left(n+1\right)n\left(n+2\right)\)
\(S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Ta có:
k(k+1)(k+2)-(k-1)k(k+1)=k.(k+1).[(k+2)-(k-1)]
=k.(k+1)(k+2-k+1)
=3k.(k+1)
Phần 2 đề sai phải là tính S=1.2.3+2.3.4+...+n.(n+1).(n+2)
