A) \[{{S}_{n}}=\frac{n{{(n+1)}^{2}}}{2},\] If n is even
B) \[{{S}_{n}}=\frac{{{n}^{2}}(n+1)}{2},\] If n is odd
C) Both and are true
D) Both and are false
Correct Answer: C
Solution :
| [c] Let \[P(n):{{S}_{n}}\] |
| \[=\left\{ \begin{matrix} \frac{n{{(n+1)}^{2}}}{2},\,\,\,\,\,when\text{ }n\text{ }is\text{ }even \\ \frac{{{n}^{2}}(n+1)}{2},\,\,\,\,\,\,when\text{ }n\text{ }is\text{ }odd \\ \end{matrix} \right.\] |
| Also, note that any term \[{{T}_{n}}\] of the series is given by |
| \[{{T}_{n}}=\left\{ \begin{matrix} {{n}^{2}},\,\,\,\,\,\,if\text{ }n\text{ }is\text{ }odd \\ 2{{n}^{2}},\,\,\,\,\,if\text{ }n\text{ }is\text{ }even \\ \end{matrix} \right.\] |
| We observe that P(1) is true, since |
| \[P(1):{{S}_{1}}={{1}^{2}}=1=\frac{1.2}{2}=\frac{{{1}^{2}}.(1+1)}{2}\] |
| Assume that P (K) is true for some natural number k, i.e |
| Case I: When k is odd, then \[k+1\] is even. We have, |
| \[P(k+1):{{S}_{k+1}}={{1}^{2}}+2\times {{2}^{2}}+...+{{k}^{2}}+2\times {{(k+1)}^{2}}\] |
| \[=\frac{{{k}^{2}}(k+1)}{2}+2\times {{(k+1)}^{2}}\] |
| \[=\frac{(k+1)}{2}[{{k}^{2}}+4(k+1)]=\frac{k+1}{2}[{{k}^{2}}+4k+4]\] |
| \[=\frac{k+1}{2}{{(k+2)}^{2}}=(k+1)\frac{{{[(k+1)+1]}^{2}}}{2}\] |
| So, \[P(k+1)\] is true, whenever P (k) is true. In the case when k is odd. |
| Case II: When k is even, then k + 1 is odd |
| Now, \[P(k+1):{{S}_{k+1}}={{1}^{2}}+2\times {{2}^{2}}+...+2.{{k}^{2}}+{{(k+1)}^{2}}\] |
| \[=\frac{k{{(k+1)}^{2}}}{2}={{(k+1)}^{2}}=\frac{{{(k+1)}^{2}}((k+1)+1)}{2}\] |
| Therefore, \[P(k+1)\] is true, whenever P (k) is true for the case when k is even. |
| Thus, P (k+1) is true whenever P (k) is true for any natural number k. Hence, P (n) true for all natural numbers. N. |
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