C0-a-a2-C1-2-a3-C2-3-an-1-Cn-n-1-

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Formation of Algebraic Expressions

C0.a+a2.C1/2+...

Question

$C_{0} . a + \frac{a^{2} . C_{1}}{2} + \frac{a^{3} . C_{2}}{3} + \dots \dots + \frac{a^{n + 1} . C_{n}}{n + 1} =$

\frac{(1 + a)^{n + 1} - 1}{n + 1}

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\frac{(1 + a)^{n + 1} - 1}{a (n + 1)}

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\frac{(1 + a)^{n} - 1}{n + 1}

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\frac{(1 + a)^{n} - 1}{a (n + 1)}

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Solution

The correct option is A $\frac{(1 + a)^{n + 1} - 1}{n + 1}$
$(1 + a)^{n} = 1 + a^{n} C_{1} + a^{2}^{n} C_{2} . . . + a^{n}^{n} C_{n}$
Integrating both the sides with respect to $a$ , we get
$\frac{(1 + a)^{n + 1}}{n + 1} + C = a + \frac{a^{2}^{n} C_{1}}{2} + \frac{a^{3}^{n} C_{2}}{3} . . . + \frac{a^{n + 1}^{n} C_{n}}{n + 1}$
Now RHS is zero for $a = 0$
Hence
$a + \frac{a^{2}^{n} C_{1}}{2} + \frac{a^{3}^{n} C_{2}}{3} . . . + \frac{a^{n + 1}^{n} C_{n}}{n + 1} = \frac{(1 + a)^{n + 1} - 1}{n + 1}$

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Similar questions

Q. Let

A_{n} = {^{n} C}_{0} {^{n} C}_{1} + {^{n} C}_{1} {^{n} C}_{2} + . . . . + {^{n} C}_{n - 1} {^{n} C}_{n}

and

\frac{A_{n + 1}}{A_{n}} = \frac{15}{4}

, then

n

equals

Q. The variance of the data:

x:	$1$	$a$	$a^{2}$	......	$a^{n}$
f:	${^{n} C}_{0}$	${^{n} C}_{1}$	${^{n} C}_{2}$	......	${^{n} C}_{n}$

a_{1}

a_{2}

a_{3}

,.....,

a_{n}

are in an

A . P .

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{(n - 1)}{\sqrt{a_{n}} + \sqrt{a_{1}}}

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

Q. If

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

denote the coefficients in the expansion of

(1 + x)^{n}

, prove that

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

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C0.a+a2.C12+a3.C23+……+an+1.Cnn+1=

The correct option is A (1+a)n+1−1n+1(1+a)n=1+anC1+a2nC2...+annCnIntegrating both the sides with respect to a, we get(1+a)n+1n+1+C=a+a2nC12+a3nC23...+an+1nCnn+1Now RHS is zero for a=0Hencea+a2nC12+a3nC23...+an+1nCnn+1=(1+a)n+1−1n+1

$C_{0} . a + \frac{a^{2} . C_{1}}{2} + \frac{a^{3} . C_{2}}{3} + \dots \dots + \frac{a^{n + 1} . C_{n}}{n + 1} =$