If n is a positive integer, then _{r=1}ⁿ r² · C_r = ( __________ ) 2ⁿ-2}

Correct Answer is : n(n+1)n(n+1)n(n+1)

Correct Answer is : n(n+1)n(n+1)n(n+1)

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AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 29 of 160

Marks: +1, -0

n

\sum\limits_{r=1}^{n} r^2 \cdot C_r

2^{n-2}

Solution:

(c)

\sum\limits_{r=1}^{n} r^{2} C_{r}=1^{2} C_{1}+2^{2} C_{2}+3^{2} C_{3}+\cdots +n^{2} C_{n}

\because (1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\cdots +C_{n} x^{n}

On differentiating both sides w.r.t.

x

, we get

n(1+x)^{n-1}=1 \cdot C_{1}+2 C_{2} x+3 C_{3} x^{2}+\cdots +n C_{n} x^{n-1}

Now, on multiplying by

x

both sides, we get

n x(1+x)^{n-1}=1 \cdot C_{1} x+2 C_{2} x^{2}+3 C_{3} x^{3}+\cdots +n C_{n} x^{n}

Now again on differentiating both sides w.r.t

x

, we get

n[(1\;+x)^{n-1}+(n-1) x(1+x)^{n-2}]

\;\; =1^{2} C_{1}+2^{2} C_{2} x+3^{2} C_{3} x^{2}+\cdots +n^{2} C_{n} x^{n-1}

Put

x=1

, we get

1^{2} \cdot C_{1}+2^{2} C_{2}\; +3^{2} C_{3}+\cdots +n^{2} C_{n}

\;\; =n[2^{n-1}+(n-1) 2^{n-2}]

\;\; =n 2^{n-2}[2+n-1]=n(n+1) 2^{n-2}

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