A Formula For The Sum Of The Triangular Numbers?
Answer : Formula for T(n): T ( n ) = n ( n + 1 ) ( n + 2 ) 6 T(n)=\frac{n(n+1)(n+2)}{6} T ( n ) = 6 n ( n + 1 ) ( n + 2 ) Proof: we can prove it in an inductive way. Base case: T ( 1 ) = 1 ∗ 2 ∗ 3 6 = 1 T(1)=\frac{1*2*3}{6}=1 T ( 1 ) = 6 1 ∗ 2 ∗ 3 = 1 Let n=k. We have T ( n + 1 ) = T ( n ) + t ( n + 1 ) . T(n+1)=T(n)+t(n+1). T ( n + 1 ) = T ( n ) + t ( n + 1 ) . Therefore T ( k + 1 ) = T ( k ) + t ( k + 1 ) = k ( k + 1 ) ( k + 2 ) 6 + ( k + 1 ) ( k + 2 ) 2 = k ( k + 1 ) ( k + 2 ) + 3 ( k + 1 ) ( k + 2 ) 6 = ( k + 1 ) ( k + 2 ) ( k + 3 ) 6 T(k+1)=T(k)+t(k+1)=\frac{k(k+1)(k+2)}{6}+\frac{(k+1)(k+2)}{2}=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{6}=\frac{(k+1)(k+2)(k+3)}{6} T ( k + 1 ) = T ( k ) + t ( k + 1 ) = 6 k ( k + 1 ) ( k + 2 ) + 2 ( k + 1 ) ( k + 2 ) = 6 k ( k + 1 ) ( k + 2 ) + 3 ( k + 1 ) ( k + 2 ) = 6 ( k + 1 ) ( k + 2 ) ( k + 3 ) Q.E.D n t h nth n t h triangular number is the sum of n n n consecutive natural numbers from starting which is simply n ( n + 1 ) / 2 n(n...