f.4 mathematical induction

Solution

Let P (n) be the proposition 1^2+2^2+3^2+...+n^2= (1/6) n (n+1) (2n+1),
where n is a natural number

(i) When n = 1
L H S =1^2 = 1
R H S = (1/6)(1) (1 + 1) [2(1) + 1] = (1/6)(1)(2)(3) = 1
L H S = R H S
∴P (1) is true.

(ii) Assume P (k) is true, where k is a natural number
That is, 1^2+2^2+3^2+...+k^2= (1/6) k (k+1) (2k+1)

L H S of P (k+1)
= 1^2+2^2+3^2+...+k^2 + (k+1)^2
= (1/6) k (k+1) (2k+1) + (k+1)^2
= (1/6) [k (k+1) (2k+1)] + 6[(k+1)^2] / 6
= (1/6) (k+1) [k (2k+1) + 6(k+1)]
= (1/6) (k+1) [2k^2 + k + 6k + 6]
= (1/6) (k+1) (2k^2 + 7k + 6)
= (1/6) (k+1) (k + 2) (2k + 3)
= (1/6) (k+1) [(k + 1) + 1] (2k + 2 + 1)
= (1/6) (k+1) [(k + 1) + 1] [2(k + 1) + 1]
= R H S of P (k+1)
∴ P (k + 1) is true if P (k) is true

Combine (i) and (ii), by the principle of Mathematical Induction, P (n) is true for all natural numbers n.

1^2 + 3^2 + 5^2 +...+ 47^2 + 49^2
= (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 +...+ 47^2 48 ^2 + 49^2 + 50^2) – (2^2 + 4^2 + 6^2 + … + 48^2 + 50^2)
= (1/6) (50) (50+1) [2(50) +1] - (2^2 + 4^2 + 6^2 + … + 48^2 + 50^2)
= (1/6) (50) (50+1) [2(50) +1] – {[(2) (1)] ^2 + [(2) (2)] ^2 +[(2) (3)] ^2 +…+ [(2) (25)] ^2}
= (1/6) (50) (50+1) [2(50) +1] – [4(1) ^2 + 4 (2) ^2 +4(3) ^2 +…+ 4 (25) ^2]
= (1/6) (50) (50+1) [2(50) +1] – 4[(1) ^2 + (2) ^2 + (3) ^2 +…+ (25) ^2]
= (1/6) (50) (50+1) [2(50) +1] – 4(1/6) (25) (25+1) [2(25) +1]
= (1/6) (50) (51) (101) – (1/6) (100) (26) (51)
= (1/6) (50) (51) [(101) – 2 (26)]
= (1/2) (50) (17) [101 - 52]
= (25) (17) (49)
= 20825