What will be the value of $n^{4} - 10 n^{3} + 36 n^{2} - 49 n + 24$ if $n = 1$ ?

21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2$
$n^{4} - 10 n^{3} + 36 n^{2} - 49 n + 24$ if $n = 1$
$= {(1)}^{4} - 10 \times {(1)}^{3} + 36 \times {(1)}^{2} - 49 \times 1 + 24$
$= 1 - 10 + 36 - 49 + 24 = 2$

Suggest Corrections

Similar questions

m = 3

n = 2

9 m^{2} - 10 n^{3}

n

1 \cdot 2^{1} + 2 \cdot 2^{2} + 3 \cdot 2^{3} + 4 \cdot 2^{4} + \dots + n \cdot 2^{n} = 2 + 2^{n + 5},

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \dots + \frac{1}{{(n + 1)}^{2}} = \frac{1}{1 -} \frac{1^{4}}{1^{2} + 2^{2} -} \dots \frac{n^{4}}{n^{2} + {(n + 1)}^{2}}

\forall n \in N

\frac{n^{4}}{24} + \frac{n^{3}}{4} + \frac{11 n^{2}}{24} + \frac{n}{4}

lim n \to \infty \frac{(1^{2} + 2^{2} + . . . . . . . + n^{2}) (1^{4} + 2^{4} + . . . . . . . + n^{4})}{(1^{7} + 2^{7} + . . . . + n^{7})} = \frac{K + 1}{15};

Join BYJU'S Learning Program