Exams Joint Entrance Examination Mathematics Binomial Theorem Question 12 Question 12 of 347 Medium MCQ
Let C r {C}_{r} C r denote the coefficient of x r {x}^{r} x r in the binomial expansion of ( 1 + x ) n , n ∈ N , 0 ≤ r ≤ n (1+x{)}^{n},n\in N,0\le r\le n ( 1 + x ) n , n ∈ N , 0 ≤ r ≤ n . If
P n = C 0 − C 1 + 2 2 3 C 2 − 2 3 4 C 3 + … . . + ( − 2 ) n n + 1 C n {P}_{n}={C}_{0}-{C}_{1}+\frac{{2}^{2}}{3}{C}_{2}-\frac{{2}^{3}}{4}{C}_{3}+\ldots ..+\frac{(-2{)}^{n}}{n+1}{C}_{n} P n = C 0 − C 1 + 3 2 2 C 2 − 4 2 3 C 3 + … .. + n + 1 ( − 2 ) n C n , then the value of ∑ n = 1 25 1 P 2 n \sum _{n=1}^{25}\frac{1}{{P}_{2n}} ∑ n = 1 25 P 2 n 1 equals.
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P n = ∑ r = 0 n n C r ( − 2 ) r r + 1 = ∑ r = 0 n 1 ( n + 1 ) n + 1 C r + 1 ( − 2 ) r = − 1 2 ( n + 1 ) ∑ r = 0 n n + 1 C r + 1 ( − 2 ) r + 1 = − 1 2 ( n + 1 ) [ ( 1 − 2 ) n + 1 − 1 ] P n = 1 2 ( n + 1 ) [ 1 − ( − 1 ) n + 1 ] \begin{gathered} {P}_{n} =\sum _{r=0}^{n}\frac{{}^{n}{C}_{r}(-2{)}^{r}}{r+1}=\sum _{r=0}^{n}\frac{1}{(n+1)}{}^{n+1}{C}_{r+1}(-2{)}^{r} \\ =\frac{-1}{2(n+1)}\sum _{r=0}^{n}{}^{n+1}{C}_{r+1}(-2{)}^{r+1}=\frac{-1}{2(n+1)}[(1-2{)}^{n+1}-1] \\ {P}_{n} =\frac{1}{2(n+1)}[1-(-1{)}^{n+1}] \end{gathered} P n = r = 0 ∑ n r + 1 n C r ( − 2 ) r = r = 0 ∑ n ( n + 1 ) 1 n + 1 C r + 1 ( − 2 ) r = 2 ( n + 1 ) − 1 r = 0 ∑ n n + 1 C r + 1 ( − 2 ) r + 1 = 2 ( n + 1 ) − 1 [( 1 − 2 ) n + 1 − 1 ] P n = 2 ( n + 1 ) 1 [ 1 − ( − 1 ) n + 1 ]
P 2 n = 1 2 ( 2 n + 1 ) [ 1 − ( − 1 ) 2 n + 1 ] P 2 n = 1 2 n + 1 ∑ n = 0 1 P 2 n = ∑ n = 1 25 ( 2 n + 1 ) = 3 + 5 + … . . + 51 = 25 2 [ 51 + 3 ] = 25 × 27 = 675 \begin{gathered} {P}_{2n}=\frac{1}{2(2n+1)}[1-(-1{)}^{2n+1}] \\ {P}_{2n}=\frac{1}{2n+1} \\ \sum _{n=0}\frac{1}{{P}_{2n}}=\sum _{n=1}^{25}(2n+1)=3+5+\ldots ..+51=\frac{25}{2}[51+3]=25\times 27=675 \end{gathered} P 2 n = 2 ( 2 n + 1 ) 1 [ 1 − ( − 1 ) 2 n + 1 ] P 2 n = 2 n + 1 1 n = 0 ∑ P 2 n 1 = n = 1 ∑ 25 ( 2 n + 1 ) = 3 + 5 + … .. + 51 = 2 25 [ 51 + 3 ] = 25 × 27 = 675 Answer the question to unlock