Find the sum - - n -11012 n -1-15 n -1

#8721;n=11012n 1 #160;+ #160;15n+1= #8721;n=11012n 1 #160;+ #160; #8721;n=11015n+1 #160; #160; #160;= #160;1+12+14+ #160;... #160;+12 ...

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Byju's Answer

Standard VII

Mathematics

Addition and Subtraction of Algebraic Expressions

Find the sum ...

Question

Find the sum : $\sum_{n = 1}^{10} [{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}] .$

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Solution

$\sum_{n = 1}^{10} [{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}] = \sum_{n = 1}^{10} {(\frac{1}{2})}^{n - 1} + \sum_{n = 1}^{10} {(\frac{1}{5})}^{n + 1} = \{1 + \frac{1}{2} + \frac{1}{4} + . . . + {(\frac{1}{2})}^{9}\} + \{\frac{1}{5^{2}} + \frac{1}{5^{3}} + \frac{1}{5^{4}} + . . . + \frac{1}{5^{11}}\} = 1 (\frac{1 - {(\frac{1}{2})}^{10}}{1 - \frac{1}{2}}) + \frac{1}{25} (\frac{1 - {(\frac{1}{5})}^{10}}{1 - \frac{1}{5}}) = (\frac{2^{10} - 1}{2^{9}}) + (\frac{5^{10} - 1}{4 \times 5^{11}})$

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Similar questions

Find the value of n such that $2 (4 n - 1) - 1 (5 n - 1) = n + 5$

$Solve for n:$
$2 (4 n - 1) - 1 (5 n - 1) = n + 4$

Q. For all

n \in N, 2^{4 n} - 15 n - 1

is divisibl;e by

If the sums of n terms of two arithmetic progressions are in the ratio 3n+5:5n-7, then their n^th terms are in the ratio
(a) 3n-15:n-1
(b) 3n+15:n+1
(c) 5n+13:n+1
(d) 5n-13:n-1

D_{r}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

\Rightarrow \sum_{r = 1}^{n} D_{r} =

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Find the sum : ∑n=110 12n-1+15n+1.

∑n=11012n-1 + 15n+1=∑n=11012n-1 + ∑n=11015n+1 = 1+12+14+ ... +129 + 152+153+154+ ...+1511=11-12101-12 + 1251-15101-15 =210-129 + 510-14×511

Find the sum : $\sum_{n = 1}^{10} [{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}] .$