Dear student, Total number of terms in the expansion is (2n+1)=Odd, so middle term of the exapnsion is (2n+1)+12th term=(n+1)th termNow coefficient of (n+1)th term will be an=Cn2n which occur once in the expansion Now putting x=1 in the given relation,(a0+a1+a2+ +an+ +a2n-2+a2n-1+a2n)=(1+1+12)n=3n⇒2(a0+a1+a2+ +an)=3n-an [since a0=a2n, a1=a2n-1, ,an-1=an+1 by the relation Crn=Cn-rn]⇒(a0+a1+a2+ +an)=12(3n-an)Regards

Please answer Q30

Please answer Q30 Given that (l + x + x: Show that Er(n = n 2 ( are the co-efficient). Prove that Show that the integer, immediately greater than + e N, is divisible by

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T o t a l n u m b e r o f t e r m s i n t h e e x p a n s i o n i s (2 n + 1) = O d d, s o m i d d l e t e r m o f t h e e x a p n s i o n i s \frac{(2 n + 1) + 1}{2} t h t e r m = (n + 1) t h t e r m N o w c o e f f i c i e n t o f (n + 1) t h t e r m w i l l b e a_{n} = {}^{2 n}C_{n} w h i c h o c c u r o n c e i n t h e e x p a n s i o n . N o w p u t t i n g x = 1 i n t h e g i v e n r e l a t i o n, (a_{0} + a_{1} + a_{2} + . . . + a_{n} + . . + a_{2 n - 2} + a_{2 n - 1} + a_{2 n}) = (1 + 1 + 1^{2})^{n} = 3^{n} \Rightarrow 2 (a_{0} + a_{1} + a_{2} + . . . . + a_{n}) = 3^{n} - a_{n} [\sin c e a_{0} = a_{2 n}, a_{1} = a_{2 n - 1}, . . ., a_{n - 1} = a_{n + 1} b y t h e r e l a t i o n {}^{n}C_{r} = {}^{n}C_{n - r}] \Rightarrow (a_{0} + a_{1} + a_{2} + . . . . . + a_{n}) = \frac{1}{2} (3^{n} - a_{n}) R e g a r d s

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Please answer Q30 Please answer Q30 Given that (l + x + x: Show that Er(n = n 2 ( are the co-efficient). Prove that Show that the integer, immediately greater than + e N, is divisible by

Please answer Q30

Please answer Q30 Given that (l + x + x: Show that Er(n = n 2 ( are the co-efficient). Prove that Show that the integer, immediately greater than + e N, is divisible by