A) \[p+q=r\]
B) \[p+r=q\]
C) \[p=q+r\]
D) \[p+q+r=0\]
Correct Answer: C
Solution :
Since (n+2)th term is the middle term in the expansion of \[{{(1+x)}^{2n+2}}\], therefore \[p={{\,}^{2n+2}}{{C}_{n+1}}\]. Since (n+1)th and (n+2)th terms are middle terms in the expansion of (1+x)2n+1, therefore \[q={{\,}^{2n+1}}{{C}_{n}}\] and \[r={{\,}^{2n+1}}{{C}_{n+1}}\] But \[^{2n+1}{{C}_{n}}+{{\,}^{2n+1}}{{C}_{n+1}}={{\,}^{2n+2}}{{C}_{n+1}}\] \[\therefore \,\,\,q+r=p\]
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