If total number of runs scored in matches is \\ where , and the runs scored in the match are given by , where . Find...

aligned& S_n=_k=1^n k 2^n+1-k \\& =1 2^n+2 2^n-1+3 2^n-2+ +(n-1) 2^2+n(2) \\& 12 S_n=1 2^n-1+2 2^n-2+3 2^n-3+ +(n-1) 2+naligned ,aligned& 12 S_n=1 2^n+1 2^n-1+1 2^n-2+ +2-n=2(2^n-1)2-1-n \\& S_n=2(2^n+1-2-n)aligned (given)aligned& n+14(2^n+1-2-n)=2(2^n+1-2-n) \\& n+14=2 n=7aligned...

If total number of runs scored in n matches is ( \frac{n + 1}{4} ) ( 2^{n + 1} - n - 2 ) where n > 1, and the runs scored in the k^{{th}{~}} match are given by k{.2}^{n + 1 - k}, where 1 \leq k \leq n. Find n: Sequence & Series (Mathematics)

If total number of runs scored in

n

matches is\n

\left( \frac{n + 1}{4} \right)

\\\n

\left( 2^{n + 1} - n - 2 \right)

where

n > 1

, and the runs\n scored in the

k^{\text{th}\text{~}}

match are given by\n

k{.2}^{n + 1 - k}

, where

1 \leq k \leq n

. Find

n

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

If a_{n} = \frac{3}{4} - ( \frac{3}{4} )^{2} + ( \frac{3}{4} )^{3} + ( - 1)^{n - 1}( \frac{3}{4} )^{n} and b_{n} = 1 - a_{n}, then find the least natural number n_{0} such that b_{n} > a_{n}\forall n ? n_{0}

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

Let A_{1},G_{1} and H_{1} denote the arithmetic, geometric and harmonic means . respectively of two distinct positive numbers. For n ? 2 let A_{n - 1} and H_{n - 1} have . arithmetic, geometric and harmonic means as A_{n},G_{n},H_{n} respectively. Which of the following statements is correct?

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

Different A.P. are constructed with the first term 100 , the last term 199, and integral common differences. The sum of the common differences of all such A.P. having at least 3 terms and at most 33 terms is

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

If a_1, a_2, a_n are in A.P. where a_i>0 for all i show that &\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}&=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}

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