Binomial Theorem And Its Simple Applications Question 97
Question: For given series: $ 1^{2}+2\times 2^{2}+3^{2}+2\times 4^{2}+5^{2}+2\times 6^{2}+…, $ If $ S_{n} $ is the sum of n terms, then
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Let $ P(n):S_{n} $
$ = \begin{cases} \frac{n{{(n+1)}^{2}}}{2},,whenniseven \\ \frac{n^{2}(n+1)}{2},whennisodd \\ \end{cases} . $
Also, note that any term $ T_{n} $ of the series is given by
$ T_{n}= \begin{cases} n^{2},ifnisodd \\ 2n^{2},,ifniseven \\ \end{cases} . $
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.
BETA
