Question

5. (i) Prove that 12 + 32 + 52 + ... + (2n + 1)2 = (n+1) 3 (2n + 1)(2n + 3), whenever n is a nonnegative integer.
(ii) Prove that 3 + 3.5 + 3.52 + ... + 3.5n = 3(5n+1 - 1) / 4, whenever n is a nonnegative integer.

Answer 1

i) $1^2+3^2+5^2+...+(2n+1)^2=\frac {(n+1)(2n+1)(2n+3)}{3}$

Basis step: $n=0$

$1^2=\frac {1\cdot1\cdot3}{3}=1$

Inductive Step: Assume that $n=k$

Inductive Hypothesis:

$1^2+3^2+5^2+...+(2k+1)^2=\frac {(k+1)(2k+1)(2k+3)}{3}$

Prove that

$1^2+3^2+5^2+...+(2k+1)^2+(2k+3)^2=\frac {(k+2)(2k+3)(2k+5)}{3}$

We have:

$1^2+3^2+5^2+...+(2k+1)^2+(2k+3)^2=\frac {(k+1)(2k+1)(2k+3)}{3}+(2k+3)^2=$

$=(2k+3)\frac {(k+1)(2k+1)(2k+3)+3(2k+3)}{3}=\frac {(2k+3)(2k^2+9k+10)}{3}=$

$=\frac {(2k+3)(2k+5)(2k+2)}{3}$

Therefore, the given formula is truth.

ii) $3+3\cdot5+3\cdot5^2+... +3\cdot5^n=\frac {3(5^{n+1}-1)}{4}$

Basis Step: $n=0$

$3=\frac {3(5-1)}{4}=3$

Inductive step: Assume that $n=k$

Inductive Hypothesis:

$3+3\cdot5+3\cdot5^2+... +3\cdot5^k=\frac {3(5^{k+1}-1)}{4}$

Prove that

$3+3\cdot5+3\cdot5^2+... +3\cdot5^k+3\cdot5^{k+1}=\frac {3(5^{k+2}-1)}{4}$

We have:

$3+3\cdot5+3\cdot5^2+... +3\cdot5^k+3\cdot5^{k+1}=\frac {3(5^{k+1}-1)}{4}+3\cdot5^{k+1}=$

$=\frac {3(5^{k+1}-1)+4\cdot3\cdot5^{k+1}}{4}=\frac {3(5^{k+1}-1+4\cdot5^{k+1})}{4}=$

$=\frac {3(5^{k+1}(1+4)-1)}{4}=\frac {3(5^{k+2}-1)}{4}$

Therefore, the given formula is truth.