Question 515 Marks

Show by the Principle of Mathematical induction that the sum $S_n$ of the n terms of the series $1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + ...$ is given by
$\text{S}_\text{n}=\begin{cases}\frac{\text{n}(\text{n}+1)^2}{2},\text{if n is even}\\\frac{\text{n}^2(\text{n}+1)}{2},\text{if n is odd}\end{cases}$

Answer

$S_n= 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ...$
Using induction we first show this is true for n = 2,
We get $S_2 = 1^2 + 2 \times 2^2 = 1 + 8 = 9$
From RHS, we have if n is even $\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)^2}{2}$
$\text{S}_\text{2}=\frac{2\times9}{2}=9$
Now using induction we first shiw this is true also
for n = 3, we get $S_3 = 1 + 8 + 9 = 18$
From RHS, we have if n is odd $\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)}{2}$
$\text{S}_\text{3}=\frac{9\times4}{2}=18$
Lets assume above is true for n = k, we get
$K$ is even, $S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + 2 \times k^2 ... (1)$
$K$ is odd, $S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + k^2 ... (2)$
Now lets prove for $n = k + 1$
If k is even, k + 1 is odd we get
$S_{k+1} = 1^2 + 2 \times 2^2 + 3^2 + ... + 2 \times k^2 + (k + 1)^2 ... (3)$
From above relation, we get
$S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + 2 \times k^2$ $=\frac{\text{k}(\text{k}+1)^2}{2}$
Substitute this on 3, we get
$\text{S}_{\text{k+1}}=\frac{\text{k}(\text{k}+1)^2}{2}+(\text{k}+1)^2=\frac{(\text{k}+1)^2(\text{k}+2)}{2}$
= RHS (when 'k + 1' is odd)
Hence Proved

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