F n - r -1 n - r 2 n C r - n C r -1-2 r -1 n C r - thenA- f 30-960B- f 21-483C- f 16-64D- f 11-44

The correct options are A f(30)=960 B f(21)=483f(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;] ⇒ n ...

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Byju's Answer

Standard XII

Mathematics

Combination

f n =∑ r =1 n...

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 options are
A $f (30) = 960$
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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Similar questions

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

Q. If

f (n) = α^{n} + β^{n}

and

∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, then

k

Q. If

α, β \neq 0

f (n) = α^{n} + β^{n}

and

∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = K {(1 - α)}^{2} {(1 - β)}^{2} {(α - β)}^{2}

, then

K

is equal to

Q. If

f (n) = α^{n} + β^{n}

and

∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, then k is equal to

Q. If

α, β \neq 0

, and

f (n) = α^{n} + β^{n}

and

∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = K (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, then K is equal to?

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

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