Is C1-1-C2-2-C3-3C4-4-1n-1-Cn-n -1-1-2-1-3-1-4-1-n -

The correct option is A True (1+x)^n=^n C_0+^n C_1 x+…+^n(C_n x^n). Integrate both sides (1+x)^n+1/n+1=^n C_o x+ ^nC_1 x^2/2+…+^nC_n x^n+1/n+1 Now put ...

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Byju's Answer

Standard XII

Mathematics

Integration

Is C1/1-C2/...

Question

Is $\frac{C_{1}}{1} - \frac{C_{2}}{2} + \frac{C_{3}}{3} \frac{C_{4}}{4} + . . . + (- 1)^{n - 1} . \frac{C_{n}}{n}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n}$ ?

True

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False

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Solution

The correct option is A True
$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x + \dots +^{n} (C_{n} x^{n})$
Integrate both sides
$\frac{(1 + x)^{n + 1}}{n + 1} =^{n} C_{o} x +^{n} C_{1} \frac{x^{2}}{2} + \dots + \frac{^{n} C_{n} x^{n + 1}}{n + 1}$
Now put $x$ from -1 to 0 (limits)

$\frac{1}{n + 1} = c_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n} \frac{C_{n}}{n}$
$\frac{1}{n + 1} \Rightarrow 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$

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Similar questions

\frac{C_{1}}{1} - \frac{C_{2}}{2} + \frac{C_{3}}{3} - \frac{C_{4}}{4} + . . . . . + \frac{(- 1)^{n - 1}}{n} C_{n}

= 1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} .

Q. Prove that

C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \frac{C_{3}}{4} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{(n + 1)}

Q. Prove that

\frac{C_{1}}{1} - \frac{C_{2}}{2} + \frac{C_{3}}{3} - \frac{C_{4}}{4} + . . . . . + \frac{(- 1)^{n - 1}}{n} C_{n}

= 1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} .

C_{0} + \frac{3}{2} . C_{1} + C_{2} + \frac{27}{4} . C_{3} + + \frac{3^{n}}{n + 1} . C_{n} = \frac{4^{n + 1} + 1}{3 (n + 1)}

Q. If

c_{0}, c_{1}, c_{2}, . . . . . . c_{n}

are the coefficients in the expansion

(1 + x)^{n},

where

n

is a positive integer, shew that

c_{1} - \frac{c_{2}}{2} + \frac{c_{3}}{3} - . . . . . . + \frac{(- 1)^{n - 1} c_{n}}{n} = 1 + \frac{1}{2} + \frac{1}{3} + . . . . \frac{1}{n}

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Is C11−C22+C33C44+...+(−1)n−1.Cnn =1+12+13+14+...+1n ?

The correct option is A True(1+x)n=nC0+nC1x+…+n(Cnxn)Integrate both sides(1+x)n+1n+1=nCox+nC1x22+…+nCnxn+1n+1Now put x from -1 to 0 (limits)1n+1=c0−C12+C23−…+(−1)nCnn1n+1⇒1+12+13+…+1n

Is $\frac{C_{1}}{1} - \frac{C_{2}}{2} + \frac{C_{3}}{3} \frac{C_{4}}{4} + . . . + (- 1)^{n - 1} . \frac{C_{n}}{n}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n}$ ?