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Maths-

General

Easy

Question

The value of the expression ⁿ⁺¹C₂ + 2[²C₂ + ³C₂ + ⁴C₂ + … + ⁿ C₂] is

$\frac{n (n + 1)}{2}$
$\frac{n (n + 1) (n + 2)}{6}$
$\frac{n (n + 1) (n + 2)}{3}$
none of these.

The correct answer is: $fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses over denominator 6 end fraction$

$n plus 1 C subscript 2 end subscript plus 2 open square brackets 2 C subscript 2 end subscript plus to the power of 3 end exponent C subscript 2 end subscript plus to the power of 4 end exponent C subscript 2 end subscript plus... plus to the power of n end exponent C subscript 2 end subscript close square brackets$
$equals fraction numerator open parentheses n plus 1 close parentheses n over denominator 1.2 end fraction plus 2 open square brackets fraction numerator 2.1 over denominator 2 end fraction plus fraction numerator 3.2 over denominator 2 end fraction plus fraction numerator 4.3 over denominator 2 end fraction plus..... plus fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction close square brackets equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction plus open square brackets 2.1 plus 3.2 plus 4.3 plus.... plus n open parentheses n minus 1 close parentheses close square brackets equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction plus not stretchy sum subscript r equals 2 end subscript superscript n end superscript r open parentheses r minus 1 close parentheses equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction plus not stretchy sum subscript r equals 1 end subscript superscript n end superscript r open parentheses r minus 1 close parentheses open square brackets because f o r r equals 1 comma r open parentheses r minus 1 close parentheses equals 0 close square brackets equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction plus fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction minus fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction$

Related Questions to study

1.ⁿC₁ + 2.ⁿC₂ + 3.ⁿC₃ + ….+ n.ⁿC_n is equal to

1.ⁿC₁ + 2.ⁿC₂ + 3.ⁿC₃ + ….+ n.ⁿC_n is equal to

$open square brackets 5 square root of 5 plus 11 close square brackets to the power of 2 n plus 1 end exponent$ is always

$open square brackets 5 square root of 5 plus 11 close square brackets to the power of 2 n plus 1 end exponent$ is always

A watch of meritorius students of the coaching institute is to be formed. For this a student must obtain at least 80% aggregate marks in a test of Physics, Chemistry and Math. (M.M. = 100 each).
The number of ways in which a student can qualify for the batch of meritorius students is

A watch of meritorius students of the coaching institute is to be formed. For this a student must obtain at least 80% aggregate marks in a test of Physics, Chemistry and Math. (M.M. = 100 each).
The number of ways in which a student can qualify for the batch of meritorius students is

parallel

The number of positive integers satisfying the inequality $n plus 1 C subscript n minus 2 end subscript minus to the power of n plus 1 end exponent C subscript n minus 1 end subscript less or equal than 100$ is

The number of positive integers satisfying the inequality $n plus 1 C subscript n minus 2 end subscript minus to the power of n plus 1 end exponent C subscript n minus 1 end subscript less or equal than 100$ is

If the (n + 1) numbers a, b, c, d,…….. be all different and each of them a prime number, then the number of different factors (other than 1) of a^m.b. c.d….. is

If the (n + 1) numbers a, b, c, d,…….. be all different and each of them a prime number, then the number of different factors (other than 1) of a^m.b. c.d….. is

If first, fifth and last terms of an A.P. is l, m, p respectively and sum of the A.P. is $fraction numerator open parentheses l plus p close parentheses open parentheses 4 p plus m minus s l close parentheses over denominator K open parentheses m minus l close parentheses end fraction comma text then end text K text is end text$

If first, fifth and last terms of an A.P. is l, m, p respectively and sum of the A.P. is $fraction numerator open parentheses l plus p close parentheses open parentheses 4 p plus m minus s l close parentheses over denominator K open parentheses m minus l close parentheses end fraction comma text then end text K text is end text$

parallel

The sum of n terms of the series $fraction numerator 1 to the power of 2 end exponent over denominator 1 end fraction plus fraction numerator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent over denominator 1 plus 2 end fraction plus fraction numerator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent plus 3 to the power of 2 end exponent over denominator 1 plus 2 plus 3 end fraction$ +….. is

The sum of n terms of the series $fraction numerator 1 to the power of 2 end exponent over denominator 1 end fraction plus fraction numerator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent over denominator 1 plus 2 end fraction plus fraction numerator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent plus 3 to the power of 2 end exponent over denominator 1 plus 2 plus 3 end fraction$ +….. is

If every pair from among the equations x² + px + qr = 0, x² + qx + rp = 0 and x² + rx + pq = 0 have a common root, then $open parentheses fraction numerator s u m blank o f blank r o o t s over denominator p r o d u c t blank o f blank r o o t s end fraction close parentheses$ is

If every pair from among the equations x² + px + qr = 0, x² + qx + rp = 0 and x² + rx + pq = 0 have a common root, then $open parentheses fraction numerator s u m blank o f blank r o o t s over denominator p r o d u c t blank o f blank r o o t s end fraction close parentheses$ is

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Chemistry-General

parallel

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Chemistry-General

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Chemistry-General

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Chemistry-General

parallel

HBr reacts fastest with:

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Chemistry-General

Which one of the following alcohols reacts most readily with Lucas reagent?

Which one of the following alcohols reacts most readily with Lucas reagent?

Chemistry-General

In this reaction,

the molecule (A) and the reagent (B) are :

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the molecule (A) and the reagent (B) are :

Chemistry-General

parallel

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