Fn-nr-1-r2nCr-nCr-1-2r-1nCr - then

F(n)=∑^n_r=1[r^2(^nC_r ^nC_r 1)+(2r+1)(^nC_r) ] ⇒ nbsp;f(n)=∑^n_r=1[(r^2+2r+1)^nC_r r^2· ^nC_r 1 nbsp;] ⇒ nbsp;f(n)=∑^n_r=1[(r+1)^2· ^nC_r r^2· ^nC_r 1 ...

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Byju's Answer

Standard XIII

Mathematics

Combination

fn=∑nr=1[r2nC...

Question

$f (n) = n \sum r = 1 [r^{2} (^{n} C_{r} -^{n} C_{r - 1}) + (2 r + 1) (^{n} C_{r})], then$

f (30) = 960

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f (21) = 483

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f (16) = 64

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f (11) = 44

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Solution

The correct option is B $f (21) = 483$
$f (n) = n \sum r = 1 [r^{2} (^{n} C_{r} -^{n} C_{r - 1}) + (2 r + 1) (^{n} C_{r})] \Rightarrow f (n) = n \sum r = 1 [(r^{2} + 2 r + 1)^{n} C_{r} - r^{2} \cdot^{n} C_{r - 1}] \Rightarrow f (n) = n \sum r = 1 [(r + 1)^{2} \cdot^{n} C_{r} - r^{2} \cdot^{n} C_{r - 1}] \Rightarrow f (n) = n \sum r = 1 (T_{r + 1} - T_{r}) \Rightarrow f (n) = (T_{2} - T_{1}) + (T_{3} - T_{2}) + (T_{4} - T_{3}) + \dots + (T_{n + 1} - T_{n}) \Rightarrow f (n) = (T_{n + 1} - T_{1} = (n + 1)^{2} \cdot^{n} C_{n} - 1 = (n^{2} + 2 n) ∴ f (30) = 960 and f (21) = 483.$

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f(n)=n∑r=1[r2(nCr−nCr−1)+(2r+1)(nCr)],then

$f (n) = n \sum r = 1 [r^{2} (^{n} C_{r} -^{n} C_{r - 1}) + (2 r + 1) (^{n} C_{r})], then$