Please help me on this, thanks a lot...1^2 + 2^2 + 3^2 + ... + n^2 = (n(n-1)(2n+1))/6

Question

Use mathematical induction to determine whether the statement is true or false for all natural numbers n. 1^2 + 2^2 + 3^2 + ... + n^2 = (n(n-1)(2n+1))/6

Answer 1

When n=1 the formula gives zero whereas the sum of the first term is the first term itself=1, therefore the formula is not correct. When n=2, the formula gives 10/6=5/3, whereas the sum of the first two terms is 5.

One requirement for proof by induction is that the base case is true. The base case is the result of applying the lowest natural number, that is, n=1, which as stated before is not true. So the answer is false.

S_n=an³+bn²+cn+d

(1) S₁=1=a+b+c+d

(2) S₂=5=8a+4b+2c+d

(3) S₃=14=27a+9b+3c+d

(4) S₄=30=64a+16b+4c+d

(5)=(2)-(1)=4=7a+3b+c

(6)=(3)-(2)=9=19a+5b+c

(7)=(4)-(3)=16=37a+7b+c

(8)=(6)-(5)=5=12a+2b

(9)=(7)-(6)=7=18a+2b

(10)=(9)-(8)=2=6a, a=⅓⇒b=½(5-12a)=½⇒c=4-7a-3b=4-7/3-3/2=⅙⇒d=1-(a+b+c)=0.

S_n=⅓n³+½n²+⅙n=n(2n²+3n+1)/6=n(n+1)(2n+1)/6 is the correct formula.